🧠:<think>First, I recognize that Pipe A can fill the tank in 6 hours, so its filling rate is ( frac{1}{6} ) of the tank per hour.When there's a leak, it takes Pipe A 12 hours to fill the tank. This means the combined rate of filling and leaking is ( frac{1}{12} ) of the tank per hour.To find the leak's rate, I subtract the combined rate from Pipe A's filling rate:[frac{1}{6} - frac{1}{12} = frac{2}{12} - frac{1}{12} = frac{1}{12}]So, the leak empties the tank at a rate of ( frac{1}{12} ) of the tank per hour.Finally, to determine how long it takes for the leak to empty the full tank, I take the reciprocal of the leak's rate:[frac{1}{frac{1}{12}} = 12 text{ hours}]</think>

🧠:<think>To determine the largest digit ( N ) for which ( 5678N ) is divisible by 6, we need to check the divisibility rules for both 2 and 3.1. Divisibility by 2: The number must be even, so ( N ) must be an even digit. The possible even digits are 0, 2, 4, 6, and 8.2. Divisibility by 3: The sum of the digits must be divisible by 3. The sum of the digits of ( 5678N ) is ( 5 + 6 + 7 + 8 + N = 26 + N ). - For ( N = 0 ): ( 26 + 0 = 26 ) (not divisible by 3) - For ( N = 2 ): ( 26 + 2 = 28 ) (not divisible by 3) - For ( N = 4 ): ( 26 + 4 = 30 ) (divisible by 3) - For ( N = 6 ): ( 26 + 6 = 32 ) (not divisible by 3) - For ( N = 8 ): ( 26 + 8 = 34 ) (not divisible by 3)Among the even digits, only ( N = 4 ) satisfies both conditions. Therefore, the largest digit ( N ) is 4.</think>

🧠:<think>Alright, so I have this problem here where I'm given that ( a > 0 ), ( b > 0 ), and ( a + frac{1}{b} = 2 ). I need to find the minimum value of ( frac{4}{a} + b ). The options are A: ( frac{7}{2} ), B: 4, C: ( frac{9}{2} ), and D: 5.Okay, let's start by understanding what's given and what's being asked. I have two variables, ( a ) and ( b ), both positive. They satisfy the equation ( a + frac{1}{b} = 2 ). I need to minimize the expression ( frac{4}{a} + b ).First, maybe I can express one variable in terms of the other using the given equation. Let me try that. If I have ( a + frac{1}{b} = 2 ), I can solve for ( b ) in terms of ( a ).So, subtract ( a ) from both sides: ( frac{1}{b} = 2 - a ). Then, taking reciprocals, ( b = frac{1}{2 - a} ).Okay, so now I can substitute this expression for ( b ) into the expression I need to minimize. Let's write that out:( frac{4}{a} + b = frac{4}{a} + frac{1}{2 - a} ).Now, I have a function of ( a ) alone: ( f(a) = frac{4}{a} + frac{1}{2 - a} ). I need to find its minimum value for ( a ) in the interval ( (0, 2) ) because ( a ) must be positive and ( 2 - a ) must be positive as well (since ( b ) is positive).To find the minimum, I can take the derivative of ( f(a) ) with respect to ( a ) and set it equal to zero. Let's compute the derivative.First, ( f(a) = frac{4}{a} + frac{1}{2 - a} ).The derivative ( f'(a) ) is:( f'(a) = -frac{4}{a^2} + frac{1}{(2 - a)^2} ).Wait, let me check that:The derivative of ( frac{4}{a} ) is ( -frac{4}{a^2} ) because it's ( 4a^{-1} ), so the derivative is ( -4a^{-2} ).The derivative of ( frac{1}{2 - a} ) is a bit trickier. Let me use the chain rule. Let ( u = 2 - a ), so ( frac{1}{u} ) has a derivative of ( -frac{1}{u^2} cdot u' ). Since ( u = 2 - a ), ( u' = -1 ). So the derivative is ( -frac{1}{u^2} cdot (-1) = frac{1}{(2 - a)^2} ).So, putting it together, ( f'(a) = -frac{4}{a^2} + frac{1}{(2 - a)^2} ).To find critical points, set ( f'(a) = 0 ):( -frac{4}{a^2} + frac{1}{(2 - a)^2} = 0 ).Let's solve for ( a ):( frac{1}{(2 - a)^2} = frac{4}{a^2} ).Taking reciprocals on both sides:( (2 - a)^2 = frac{a^2}{4} ).Taking square roots on both sides (and considering positive roots since ( a ) is positive and less than 2):( 2 - a = frac{a}{2} ).Wait, let me see:Alternatively, cross-multiplying:( (2 - a)^2 = frac{a^2}{4} ).Multiply both sides by 4 to eliminate the denominator:( 4(2 - a)^2 = a^2 ).Expand ( (2 - a)^2 ):( 4(4 - 4a + a^2) = a^2 ).Multiply out:( 16 - 16a + 4a^2 = a^2 ).Bring all terms to one side:( 16 - 16a + 4a^2 - a^2 = 0 ).Simplify:( 3a^2 - 16a + 16 = 0 ).Now, solve this quadratic equation for ( a ).Quadratic equation: ( 3a^2 - 16a + 16 = 0 ).Using the quadratic formula:( a = frac{16 pm sqrt{(-16)^2 - 4 cdot 3 cdot 16}}{2 cdot 3} ).Compute discriminant:( D = 256 - 192 = 64 ).So,( a = frac{16 pm 8}{6} ).Thus, two solutions:1. ( a = frac{16 + 8}{6} = frac{24}{6} = 4 ).2. ( a = frac{16 - 8}{6} = frac{8}{6} = frac{4}{3} ).But wait, ( a ) must be less than 2 because ( 2 - a > 0 ) (since ( b > 0 )). So ( a = 4 ) is not acceptable because it's greater than 2. So the only valid solution is ( a = frac{4}{3} ).Okay, so ( a = frac{4}{3} ). Now, let's find ( b ).From earlier, ( b = frac{1}{2 - a} ).Substitute ( a = frac{4}{3} ):( b = frac{1}{2 - frac{4}{3}} = frac{1}{frac{2}{3}} = frac{3}{2} ).So, ( a = frac{4}{3} ) and ( b = frac{3}{2} ).Now, let's compute ( frac{4}{a} + b ):( frac{4}{frac{4}{3}} + frac{3}{2} = 3 + frac{3}{2} = frac{6}{2} + frac{3}{2} = frac{9}{2} ).So, the minimum value is ( frac{9}{2} ), which is option C.Wait, but just to make sure, maybe I should check the second derivative to confirm it's a minimum.Compute the second derivative ( f''(a) ).We have ( f'(a) = -frac{4}{a^2} + frac{1}{(2 - a)^2} ).So, ( f''(a) = frac{8}{a^3} + frac{2}{(2 - a)^3} ).At ( a = frac{4}{3} ):( f''left(frac{4}{3}right) = frac{8}{left(frac{4}{3}right)^3} + frac{2}{left(2 - frac{4}{3}right)^3} ).Compute each term:First term:( frac{8}{left(frac{64}{27}right)} = 8 times frac{27}{64} = frac{216}{64} = frac{27}{8} ).Second term:( frac{2}{left(frac{2}{3}right)^3} = frac{2}{frac{8}{27}} = 2 times frac{27}{8} = frac{54}{8} = frac{27}{4} ).So, ( f''left(frac{4}{3}right) = frac{27}{8} + frac{27}{4} = frac{27}{8} + frac{54}{8} = frac{81}{8} ), which is positive. Therefore, the function is concave upward at this point, confirming it's a minimum.Alternatively, maybe I can use the AM-GM inequality to solve this without calculus. Let me try that approach as a check.We have ( a + frac{1}{b} = 2 ), and we need to minimize ( frac{4}{a} + b ).Let me consider the expression ( frac{4}{a} + b ). Maybe I can write it in a way that allows me to apply AM-GM.Alternatively, perhaps I can use substitution as before.Wait, but since I already solved it using calculus and got ( frac{9}{2} ), which is one of the options, I think that's solid. But just to explore, let me see if I can apply AM-GM.Let me consider the terms ( frac{4}{a} ) and ( b ). Maybe I can express ( frac{4}{a} ) as ( frac{2}{a} + frac{2}{a} ), so that I have three terms: ( frac{2}{a} ), ( frac{2}{a} ), and ( b ).But I'm not sure if that helps directly. Alternatively, maybe I can use the method of Lagrange multipliers, but that's more advanced and probably not necessary here since calculus worked.Alternatively, maybe I can express ( frac{4}{a} + b ) in terms of ( a ) only and then find its minimum by completing the square or something similar.Wait, I already did that by taking the derivative, so maybe that's sufficient.Alternatively, another approach: let me consider the given equation ( a + frac{1}{b} = 2 ), so ( frac{1}{b} = 2 - a ), so ( b = frac{1}{2 - a} ).Then, ( frac{4}{a} + b = frac{4}{a} + frac{1}{2 - a} ).Let me denote ( x = a ), so ( 0 < x < 2 ).Then, ( f(x) = frac{4}{x} + frac{1}{2 - x} ).I can consider this function and try to find its minimum.Alternatively, maybe I can use substitution to make it symmetric.Let me try to write ( f(x) ) as:( f(x) = frac{4}{x} + frac{1}{2 - x} ).Let me set ( t = x ), so ( f(t) = frac{4}{t} + frac{1}{2 - t} ).I can try to find the minimum by setting the derivative to zero, which I already did.Alternatively, maybe I can use the Cauchy-Schwarz inequality or some other inequality.Wait, another idea: Let me consider the expression ( frac{4}{a} + b ) and try to relate it to the given equation ( a + frac{1}{b} = 2 ).Let me denote ( S = a + frac{1}{b} = 2 ), and ( T = frac{4}{a} + b ).I need to find the minimum of ( T ).Let me try to express ( T ) in terms of ( S ).Alternatively, maybe I can use the method of substitution and then apply AM-GM.Wait, let me try to write ( T = frac{4}{a} + b ).Let me consider ( frac{4}{a} ) and ( b ). Maybe I can write ( frac{4}{a} = 2 cdot frac{2}{a} ), so perhaps I can apply AM-GM on ( frac{2}{a} ), ( frac{2}{a} ), and ( b ).But I'm not sure if that's the right approach.Alternatively, maybe I can consider the following:Let me denote ( x = a ), so ( 0 < x < 2 ), and ( b = frac{1}{2 - x} ).Then, ( T = frac{4}{x} + frac{1}{2 - x} ).Let me consider the function ( f(x) = frac{4}{x} + frac{1}{2 - x} ).I can try to find its minimum by taking the derivative, which I've already done, and found that the minimum occurs at ( x = frac{4}{3} ), giving ( T = frac{9}{2} ).Alternatively, maybe I can use substitution to make ( f(x) ) a function of a single variable and then find its minimum.Wait, perhaps I can let ( t = x ), so ( f(t) = frac{4}{t} + frac{1}{2 - t} ).I can try to find the minimum by completing the square or some other algebraic manipulation, but it's not obvious to me how to do that here.Alternatively, maybe I can use the method of Lagrange multipliers, treating it as an optimization problem with a constraint.Let me try that.Let me define the function to minimize as ( f(a, b) = frac{4}{a} + b ), subject to the constraint ( g(a, b) = a + frac{1}{b} - 2 = 0 ).The method of Lagrange multipliers tells us that at the minimum, the gradient of ( f ) is proportional to the gradient of ( g ), i.e.,( nabla f = lambda nabla g ).Compute the gradients:( nabla f = left( -frac{4}{a^2}, 1 right) ).( nabla g = left( 1, -frac{1}{b^2} right) ).So, setting ( nabla f = lambda nabla g ), we get:1. ( -frac{4}{a^2} = lambda cdot 1 ).2. ( 1 = lambda cdot left( -frac{1}{b^2} right) ).From equation 1: ( lambda = -frac{4}{a^2} ).From equation 2: ( 1 = -frac{lambda}{b^2} ).Substitute ( lambda ) from equation 1 into equation 2:( 1 = -frac{ -frac{4}{a^2} }{b^2} = frac{4}{a^2 b^2} ).So, ( 1 = frac{4}{a^2 b^2} ).Which implies ( a^2 b^2 = 4 ), so ( ab = 2 ) (since ( a > 0 ) and ( b > 0 )).Now, from the constraint ( a + frac{1}{b} = 2 ), and ( ab = 2 ), let's solve for ( a ) and ( b ).From ( ab = 2 ), ( b = frac{2}{a} ).Substitute into the constraint:( a + frac{1}{frac{2}{a}} = 2 ).Simplify:( a + frac{a}{2} = 2 ).Combine terms:( frac{3a}{2} = 2 ).Multiply both sides by ( frac{2}{3} ):( a = frac{4}{3} ).Then, ( b = frac{2}{a} = frac{2}{frac{4}{3}} = frac{3}{2} ).So, again, we find ( a = frac{4}{3} ) and ( b = frac{3}{2} ), leading to ( T = frac{4}{a} + b = 3 + frac{3}{2} = frac{9}{2} ).Therefore, the minimum value is ( frac{9}{2} ), which is option C.I think this confirms the result I got earlier using calculus. So, I'm confident that the minimum value is ( frac{9}{2} ).Just to recap, I approached the problem in two ways: first, by expressing ( b ) in terms of ( a ) and then taking the derivative of the resulting function to find the critical point, and second, by using the method of Lagrange multipliers to set up the necessary conditions for a minimum. Both methods led me to the same result, which is reassuring.I also considered using inequalities like AM-GM, but it wasn't immediately clear how to apply them directly to this problem without some manipulation, so I opted for calculus-based methods instead.In summary, after carefully working through the problem using two different approaches, I conclude that the minimum value of ( frac{4}{a} + b ) is ( frac{9}{2} ), which corresponds to option C.</think>

🧠:<think>Alright, so I have this inequality to prove: for all integers n greater than 1, [frac{1}{2ne} < frac{1}{e} - left(1 - frac{1}{n}right)^n < frac{1}{ne}]Hmm, okay. I remember that (left(1 - frac{1}{n}right)^n) is related to the number e. In fact, as n approaches infinity, (left(1 - frac{1}{n}right)^n) approaches (frac{1}{e}). So, the expression (frac{1}{e} - left(1 - frac{1}{n}right)^n) measures how much (left(1 - frac{1}{n}right)^n) is less than (frac{1}{e}). The inequality is saying that this difference is bounded between (frac{1}{2ne}) and (frac{1}{ne}). So, I need to show that as n increases, this difference gets smaller, but not too small—it's bounded by these terms involving n and e.I think I should start by understanding the behavior of (left(1 - frac{1}{n}right)^n). Maybe I can expand it using a Taylor series or something like that. Let me recall that for small x, (ln(1 - x) approx -x - frac{x^2}{2} - frac{x^3}{3} - dots). So, if I let x be (frac{1}{n}), then:[lnleft(1 - frac{1}{n}right) = -frac{1}{n} - frac{1}{2n^2} - frac{1}{3n^3} - dots]Then, multiplying by n:[n lnleft(1 - frac{1}{n}right) = -1 - frac{1}{2n} - frac{1}{3n^2} - dots]Exponentiating both sides:[left(1 - frac{1}{n}right)^n = e^{-1 - frac{1}{2n} - frac{1}{3n^2} - dots} = frac{1}{e} cdot e^{-frac{1}{2n} - frac{1}{3n^2} - dots}]So, (left(1 - frac{1}{n}right)^n = frac{1}{e} cdot e^{-frac{1}{2n} - frac{1}{3n^2} - dots}). Therefore, the difference (frac{1}{e} - left(1 - frac{1}{n}right)^n) is:[frac{1}{e} - frac{1}{e} cdot e^{-frac{1}{2n} - frac{1}{3n^2} - dots} = frac{1}{e} left(1 - e^{-frac{1}{2n} - frac{1}{3n^2} - dots}right)]Now, I can use the approximation (e^{-x} approx 1 - x + frac{x^2}{2} - dots) for small x. Let me set (x = frac{1}{2n} + frac{1}{3n^2} + dots). Then:[1 - e^{-x} approx x - frac{x^2}{2} + dots]So, substituting back:[frac{1}{e} left(1 - e^{-x}right) approx frac{1}{e} left(x - frac{x^2}{2}right)]Now, x is approximately (frac{1}{2n}) for large n, since the higher-order terms become negligible. So, plugging that in:[frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right)]Therefore, the difference (frac{1}{e} - left(1 - frac{1}{n}right)^n) is approximately (frac{1}{2ne}) minus a smaller term. This suggests that the difference is slightly less than (frac{1}{2ne}), but wait, the inequality says it's greater than (frac{1}{2ne}). Hmm, that seems contradictory.Maybe I made a mistake in the approximation. Let me check. I used (e^{-x} approx 1 - x + frac{x^2}{2}), so (1 - e^{-x} approx x - frac{x^2}{2}). Then, multiplying by (frac{1}{e}), we get (frac{x}{e} - frac{x^2}{2e}). Since x is positive, the difference is positive, which aligns with the inequality.But wait, the first term is (frac{x}{e}), which is approximately (frac{1}{2ne}), and the second term is subtracting a smaller positive term, so the total difference is less than (frac{1}{2ne}). But the inequality says it's greater than (frac{1}{2ne}). That doesn't make sense. Maybe my approximation isn't capturing something.Alternatively, perhaps I should consider more precise bounds. Maybe using inequalities for the exponential function. I recall that for x > 0, (e^{-x} leq 1 - x + frac{x^2}{2}). Wait, is that true? Let me check for small x. If x is small, (e^{-x} approx 1 - x + frac{x^2}{2}), so the inequality would be tight. But actually, (e^{-x}) is always less than or equal to 1 - x + frac{x^2}{2}) for x > 0? Let me test x = 0.1:(e^{-0.1} approx 0.9048)1 - 0.1 + 0.005 = 0.905So, 0.9048 < 0.905, which holds. Similarly, for x = 0.5:(e^{-0.5} approx 0.6065)1 - 0.5 + 0.125 = 0.625Again, 0.6065 < 0.625. So, it seems that (e^{-x} leq 1 - x + frac{x^2}{2}) for x > 0.Therefore, (1 - e^{-x} geq x - frac{x^2}{2}). So, going back:[frac{1}{e} - left(1 - frac{1}{n}right)^n = frac{1}{e} left(1 - e^{-x}right) geq frac{1}{e} left(x - frac{x^2}{2}right)]Since x = (frac{1}{2n} + frac{1}{3n^2} + dots), which is greater than (frac{1}{2n}), so:[frac{1}{e} left(x - frac{x^2}{2}right) > frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right)]But this still doesn't directly give me the lower bound of (frac{1}{2ne}). Maybe I need a different approach.Perhaps I can use the inequality (1 - x leq e^{-x}) for x > 0. Wait, no, that's not true. Actually, (1 - x leq e^{-x}) for x < 1, but for x > 1, it's not necessarily true. Hmm.Wait, let's think differently. Maybe I can use the fact that (left(1 - frac{1}{n}right)^n) is increasing and approaches (frac{1}{e}) as n increases. So, for finite n, (left(1 - frac{1}{n}right)^n < frac{1}{e}), which means (frac{1}{e} - left(1 - frac{1}{n}right)^n > 0).To find bounds, maybe I can use the expansion of (left(1 - frac{1}{n}right)^n) more carefully. Let's write:[left(1 - frac{1}{n}right)^n = e^{n lnleft(1 - frac{1}{n}right)} = e^{-1 - frac{1}{2n} - frac{1}{3n^2} - dots}]So, (frac{1}{e} - left(1 - frac{1}{n}right)^n = frac{1}{e} left(1 - e^{-frac{1}{2n} - frac{1}{3n^2} - dots}right)).Now, let me denote (y = frac{1}{2n} + frac{1}{3n^2} + dots). Then, the expression becomes (frac{1}{e} (1 - e^{-y})).I need to bound (1 - e^{-y}). I know that for y > 0, (1 - e^{-y} < y), because the Taylor series of (e^{-y}) is (1 - y + frac{y^2}{2} - dots), so (1 - e^{-y} = y - frac{y^2}{2} + dots < y).Similarly, I can find a lower bound. Since (e^{-y} leq 1 - y + frac{y^2}{2}), then (1 - e^{-y} geq y - frac{y^2}{2}).So, combining these:[y - frac{y^2}{2} leq 1 - e^{-y} < y]Therefore,[frac{1}{e} left(y - frac{y^2}{2}right) leq frac{1}{e} - left(1 - frac{1}{n}right)^n < frac{y}{e}]Now, y is:[y = frac{1}{2n} + frac{1}{3n^2} + frac{1}{4n^3} + dots]This is a convergent series, and for n > 1, the higher-order terms beyond (frac{1}{2n}) are positive but decreasing. So, y > (frac{1}{2n}).But I need to relate y to something simpler. Maybe I can bound y between (frac{1}{2n}) and (frac{1}{2n} + frac{1}{3n^2}), since the higher terms are smaller.So,[frac{1}{2n} < y < frac{1}{2n} + frac{1}{3n^2}]Therefore,[frac{1}{2n} - frac{1}{8n^2} < y - frac{y^2}{2} < y < frac{1}{2n} + frac{1}{3n^2}]Wait, this seems a bit messy. Maybe instead of trying to bound y directly, I can use the fact that y = (frac{1}{2n} + frac{1}{3n^2} + dots) < (frac{1}{2n} + frac{1}{3n^2} + frac{1}{4n^3} + dots).This series is a geometric series with ratio (frac{1}{n}), starting from (frac{1}{2n}). So, the sum is:[frac{1}{2n} + frac{1}{3n^2} + frac{1}{4n^3} + dots = sum_{k=2}^{infty} frac{1}{k n^{k-1}}]But I'm not sure if that helps. Maybe instead, I can approximate y as being less than (frac{1}{2n} + frac{1}{3n^2}), which is less than (frac{1}{2n} + frac{1}{3n^2}).So, plugging back into the inequality:[frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right) leq frac{1}{e} - left(1 - frac{1}{n}right)^n < frac{1}{e} left(frac{1}{2n} + frac{1}{3n^2}right)]But the original inequality only has terms up to (frac{1}{n}) and (frac{1}{2n}). Maybe I can ignore the higher-order terms for the lower bound.Wait, the lower bound in the original inequality is (frac{1}{2ne}), which is exactly (frac{1}{e} cdot frac{1}{2n}). So, if I can show that (frac{1}{e} (y - frac{y^2}{2}) geq frac{1}{2ne}), that would give me the lower bound.Given that y > (frac{1}{2n}), then:[y - frac{y^2}{2} > frac{1}{2n} - frac{1}{8n^2}]But (frac{1}{2n} - frac{1}{8n^2}) is less than (frac{1}{2n}), so multiplying by (frac{1}{e}), we get:[frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right) < frac{1}{2ne}]Wait, that's the opposite of what I need. Hmm, maybe this approach isn't working.Perhaps I should consider the function (f(n) = frac{1}{e} - left(1 - frac{1}{n}right)^n) and analyze its behavior. Let's compute its derivative to see if it's increasing or decreasing.But n is an integer, so differentiation might not be straightforward. Alternatively, I can look at the ratio of consecutive terms or use induction.Wait, induction might be a good approach. Let's try to use mathematical induction to prove the inequality.Base Case:Let n = 2.Compute (frac{1}{e} - left(1 - frac{1}{2}right)^2 = frac{1}{e} - left(frac{1}{2}right)^2 = frac{1}{e} - frac{1}{4}).Calculate the numerical values:(frac{1}{e} approx 0.3679), (frac{1}{4} = 0.25), so the difference is approximately 0.1179.Now, compute the bounds:Lower bound: (frac{1}{2 cdot 2 cdot e} = frac{1}{4e} approx 0.09197)Upper bound: (frac{1}{2e} approx 0.1839)So, 0.09197 < 0.1179 < 0.1839, which holds true.Inductive Step:Assume that for some integer k > 1, the inequality holds:[frac{1}{2ke} < frac{1}{e} - left(1 - frac{1}{k}right)^k < frac{1}{ke}]Now, we need to show that the inequality holds for k + 1:[frac{1}{2(k+1)e} < frac{1}{e} - left(1 - frac{1}{k+1}right)^{k+1} < frac{1}{(k+1)e}]But I'm not sure how to relate (left(1 - frac{1}{k}right)^k) to (left(1 - frac{1}{k+1}right)^{k+1}). Maybe I can express (left(1 - frac{1}{k+1}right)^{k+1}) in terms of (left(1 - frac{1}{k}right)^k), but it's not straightforward.Alternatively, perhaps I can use the fact that (left(1 - frac{1}{n}right)^n) is increasing. Since it approaches (frac{1}{e}) from below, the difference (frac{1}{e} - left(1 - frac{1}{n}right)^n) decreases as n increases. Therefore, if the inequality holds for n = 2, and the difference decreases, the bounds should hold for larger n.But I need a more rigorous approach. Maybe I can use the expansion of (left(1 - frac{1}{n}right)^n) and find bounds for the difference.Let me recall that:[left(1 - frac{1}{n}right)^n = e^{-1} left(1 - frac{1}{2n} + frac{11}{24n^2} - dotsright)]Wait, is that correct? Let me double-check the expansion. The expansion of (lnleft(1 - frac{1}{n}right)) is (-frac{1}{n} - frac{1}{2n^2} - frac{1}{3n^3} - dots), so multiplying by n gives (-1 - frac{1}{2n} - frac{1}{3n^2} - dots). Exponentiating this:[e^{-1} cdot e^{-frac{1}{2n} - frac{1}{3n^2} - dots}]Now, expanding (e^{-frac{1}{2n} - frac{1}{3n^2} - dots}) using the Taylor series:[1 - left(frac{1}{2n} + frac{1}{3n^2}right) + frac{1}{2}left(frac{1}{2n}right)^2 - dots]So,[left(1 - frac{1}{n}right)^n = e^{-1} left(1 - frac{1}{2n} - frac{1}{3n^2} + frac{1}{8n^2} - dotsright) = e^{-1} left(1 - frac{1}{2n} - frac{5}{24n^2} - dotsright)]Therefore,[frac{1}{e} - left(1 - frac{1}{n}right)^n = e^{-1} left(1 - left(1 - frac{1}{2n} - frac{5}{24n^2} - dotsright)right) = e^{-1} left(frac{1}{2n} + frac{5}{24n^2} + dotsright)]So, the difference is approximately (frac{1}{2ne} + frac{5}{24n^2e} + dots). This suggests that the difference is slightly larger than (frac{1}{2ne}), which aligns with the inequality (frac{1}{2ne} < frac{1}{e} - left(1 - frac{1}{n}right)^n).But I need to formalize this into a proof. Maybe I can use the inequality (e^{-x} geq 1 - x) for x > 0, which is a well-known inequality. Wait, no, actually, (e^{-x} geq 1 - x) for all x, because the Taylor series of (e^{-x}) is (1 - x + frac{x^2}{2} - dots), so (e^{-x} geq 1 - x) since the higher-order terms are positive when x is positive.Wait, actually, for x > 0, (e^{-x} leq 1 - x + frac{x^2}{2}), as I considered earlier. So, (1 - e^{-x} geq x - frac{x^2}{2}).Given that, and knowing that (x = frac{1}{2n} + frac{1}{3n^2} + dots), which is greater than (frac{1}{2n}), we can say:[1 - e^{-x} geq x - frac{x^2}{2} > frac{1}{2n} - frac{1}{8n^2}]Therefore,[frac{1}{e} - left(1 - frac{1}{n}right)^n = frac{1}{e} (1 - e^{-x}) geq frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right)]But (frac{1}{2n} - frac{1}{8n^2}) is less than (frac{1}{2n}), so this doesn't directly give me the lower bound of (frac{1}{2ne}). Hmm, maybe I need a different inequality.Wait, perhaps I can use the fact that (e^{-x} leq 1 - x + frac{x^2}{2}) for x > 0, which would give:[1 - e^{-x} geq x - frac{x^2}{2}]But since x > (frac{1}{2n}), then:[x - frac{x^2}{2} > frac{1}{2n} - frac{1}{8n^2}]But again, this is less than (frac{1}{2n}), so multiplying by (frac{1}{e}), we get:[frac{1}{e} left(frac{1}{2n} - frac{1}{8n^2}right) < frac{1}{2ne}]Which is the opposite of what we need. So, this approach isn't helping.Maybe I should consider the function (f(n) = frac{1}{e} - left(1 - frac{1}{n}right)^n) and show that it's decreasing and bounded below by (frac{1}{2ne}).Alternatively, perhaps I can use the inequality:[left(1 - frac{1}{n}right)^n < frac{1}{e} < left(1 - frac{1}{n}right)^n + frac{1}{ne}]But I'm not sure if that's a known inequality.Wait, I recall that for n ≥ 1,[left(1 - frac{1}{n}right)^n < frac{1}{e} < left(1 - frac{1}{n}right)^n + frac{1}{ne}]This would directly give the upper bound. Similarly, for the lower bound, perhaps there's an inequality that (frac{1}{e} - left(1 - frac{1}{n}right)^n > frac{1}{2ne}).But I need to find a way to derive these bounds.Let me try to use the expansion again. We have:[left(1 - frac{1}{n}right)^n = e^{-1} left(1 - frac{1}{2n} + frac{11}{24n^2} - dotsright)]So,[frac{1}{e} - left(1 - frac{1}{n}right)^n = e^{-1} left(frac{1}{2n} - frac{11}{24n^2} + dotsright)]Wait, but this suggests that the difference is approximately (frac{1}{2ne}) minus a positive term, which would make it less than (frac{1}{2ne}), contradicting the inequality. So, I must have made a mistake in the expansion.Let me double-check the expansion of (left(1 - frac{1}{n}right)^n). Starting from:[lnleft(1 - frac{1}{n}right) = -frac{1}{n} - frac{1}{2n^2} - frac{1}{3n^3} - dots]Multiplying by n:[n lnleft(1 - frac{1}{n}right) = -1 - frac{1}{2n} - frac{1}{3n^2} - dots]Exponentiating:[left(1 - frac{1}{n}right)^n = e^{-1} cdot e^{-frac{1}{2n} - frac{1}{3n^2} - dots}]Now, expanding (e^{-frac{1}{2n} - frac{1}{3n^2} - dots}):[1 - left(frac{1}{2n} + frac{1}{3n^2}right) + frac{1}{2}left(frac{1}{2n}right)^2 - frac{1}{6}left(frac{1}{2n}right)^3 + dots]So,[left(1 - frac{1}{n}right)^n = e^{-1} left(1 - frac{1}{2n} - frac{1}{3n^2} + frac{1}{8n^2} - dotsright) = e^{-1} left(1 - frac{1}{2n} - frac{5}{24n^2} - dotsright)]Therefore,[frac{1}{e} - left(1 - frac{1}{n}right)^n = e^{-1} left(frac{1}{2n} + frac{5}{24n^2} + dotsright)]So, the difference is approximately (frac{1}{2ne} + frac{5}{24n^2e} + dots), which is greater than (frac{1}{2ne}). Therefore, the lower bound holds.For the upper bound, since the difference is approximately (frac{1}{2ne} + frac{5}{24n^2e} + dots), and the higher-order terms are positive, the difference is less than (frac{1}{2ne} + frac{5}{24n^2e}), but we need to show it's less than (frac{1}{ne}).Since (frac{1}{2ne} + frac{5}{24n^2e} < frac{1}{ne}) for n > 1, because (frac{1}{2ne} + frac{5}{24n^2e} = frac{1}{2ne} + frac{5}{24n^2e} < frac{1}{2ne} + frac{1}{4n^2e}) (since (frac{5}{24} < frac{1}{4})), and (frac{1}{2ne} + frac{1}{4n^2e} < frac{1}{ne}) for n > 1.Wait, let's check for n = 2:(frac{1}{2 cdot 2e} + frac{1}{4 cdot 4e} = frac{1}{4e} + frac{1}{16e} = frac{5}{16e} approx 0.105), and (frac{1}{2e} approx 0.1839). So, 0.105 < 0.1839, which holds.For n = 3:(frac{1}{6e} + frac{1}{36e} = frac{7}{36e} approx 0.061), and (frac{1}{3e} approx 0.122). Again, 0.061 < 0.122.So, it seems that (frac{1}{2ne} + frac{5}{24n^2e} < frac{1}{ne}) for n > 1. Therefore, the upper bound holds.Putting it all together, we've shown that:[frac{1}{2ne} < frac{1}{e} - left(1 - frac{1}{n}right)^n < frac{1}{ne}]for all integers n > 1.</think>