Consider the sequence $(\frac{n}{n+1})^n$. I am pretty sure it monotonically decreases. But how can we prove it?
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@lulu - Could you elaborate? Mine is not the same and mine is monotonically decreasing. – mark Feb 08 '24 at 19:19
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Note that $1+\frac 1x=\frac {x+1}x$ is the reciprocal of $\frac x{x+1}$. And note that for a positive sequence, ${a_n}$ is decreasing if and only if $\frac 1{a_n}$ is increasing. – lulu Feb 08 '24 at 19:21
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@lulu - you are correct. Now I can see it. – mark Feb 08 '24 at 19:29
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Let $(x_n)_{n\in \mathbb{N}} \text{ be a sequence, where }$ $x_n=(\frac{n}{n+1})^n$.Observe that $x_n=\frac{1}{y_n}$, where $y_n=(\frac{n+1}{n})^n=(1+\frac{1}{n})^n$. It is proved that the sequence $(y_n)_{n\in \mathbb{N}^{\star}}$ is stricly increasing to $e$. As this sequence is strictly increasing, the reciprocal of it, $x_n$ is strictly decreasing to $\frac{1}{e}$($\lim_{n\to \infty}{x_n}=\lim_{n\to \infty}{\frac{1}{y_n}}=\frac{1}{e})$.
Sebastiano
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I have changed the solution and I am sure that this one is correct. – mrvalentyn Feb 08 '24 at 19:50
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Please do not answer duplicates or questions without context and where the OP does not show effort. It goes against the regulations of this site. – Gary Feb 09 '24 at 01:35