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    Hi, can someone explain to me why  (\frac{k}{k+1})^k \rightarrow \frac{1}{e} as  k \rightarrow \infty. I always thought it was 1 since  \frac{k}{k+1} \rightarrow 1 as  k \rightarrow \infty but now i know that's wrong. But I still don't know why the answer is  \frac{1}{e} . Thanks.
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    There's no limit, we'lll reach for the sky. No valley too deep, nor mountain too high.

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    (Original post by JBKProductions)
    Hi, can someone explain to me why  (\frac{k}{k+1})^k \rightarrow \frac{1}{e} as  k \rightarrow \infty. I always thought it was 1 since  \frac{k}{k+1} \rightarrow 1 as  k \rightarrow \infty but now i know that's wrong. But I still don't know why the answer is  \frac{1}{e} . Thanks.
    You can write it as \dfrac{k+1}{k} \left( \dfrac{k}{k+1} \right)^{k+1}, and then split the fraction inside the bracket to give \dfrac{k+1}{k} \left( 1 - \dfrac{1}{k+1} \right)^{k+1}. Then you can use the fact that \left( 1 + \dfrac{x}{n} \right)^n \to e^x as n \to \infty.

    EDIT: I should add that you can see that the last part is true using a binomial expansion.
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    Thanks!
 
 
 

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