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The problem is

Suppose that $\{x_n\}$ is positive, monotonic decreasing and $\sum_{n=1}^\infty x_n=+\infty$. Prove that $$\sum_{n=1}^\infty x_ne^{-\frac{x_{n}}{x_{n+1}}}=+\infty.$$

This is a past examination problem of Analysis. I tried many ways, but failed. One trival idea is to prove $\frac{x_{n}}{x_{n+1}}$ is bounded above, but I failed to show it, maybe it is not ture. Hope to find some hints here, thanks in advance.

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    $\begingroup$ By ratio test, we know $\limsup \frac{x_{n+1}}{x_n}$ is not less than $1$ (decreasing conditions and positivity of infinite sum means all terms must be positive, so we can drop absolute values), so in particular the inverse quotient is not eventually greater than $1$, so $\exp(-\frac{x_{n}}{x_{n+1}} )$ is not eventually less than $1/e$ so the modified sum should eventually be roughly at least as big as $1/e$ times the original sum (and thus diverges). There's some fanagling to be done to formalize that idea, but it should suffice. $\endgroup$ Apr 4, 2023 at 8:49
  • $\begingroup$ Could you give more details on what you tried? $\endgroup$
    – gist076923
    Apr 4, 2023 at 13:08

2 Answers 2

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Clearly, the sequence $\{x_n\}$ converges. If $x_n\to a>0$, then $x_n/x_{n+1}\to 1$, and hence $\mathrm{e}^{-x_n/x_{n+1}}$ is lower bounded, say by $b>0$, and hence $$ \sum x_n \mathrm{e}^{-x_n/x_{n+1}}\ge b\sum x_n =\infty. $$ If $x_n\to 0$, set $$ S_k=\{n\in\mathbb N: 2^{-k}<x_n\le 2^{-k+1}\} $$ Then $\sum_{k=0}^\infty\sum_{n\in S_k}x_n=\infty$, and $$ \sum_{k=0}^\infty2^{-k+1}|S_k|\ge\sum_{k=0}^\infty\sum_{n\in S_k}x_n\ge \sum_{k=0}^\infty 2^{-k}|S_k|. $$ Therefore $$\sum_{k=0}^\infty2^{-k}|S_k|=\frac{1}{2}\sum_{k=0}^\infty2^{-k+1} |S_k|=\infty.$$ Also, if $n,n+1\in S_k$, then $\mathrm{e}^{-x_n/x_{n+1}}\ge \mathrm{e}^{-2}$. Hence, if $S_k=\{j,j+1,\ldots,\ell\}$, then $$ \sum_{n\in S_k}x_n\mathrm{e}^{-x_n/x_{n+1}}>\sum_{n=j}^{\ell-1} x_n\mathrm{e}^{-x_n/x_{n+1}}\ge \sum_{n=j}^{\ell-1} x_n\mathrm{e}^{-2}\ge 2^{-k}(|S_k|-1)\mathrm{e}^{-2}. $$ Thus $$ \sum_{k=0}^\infty\sum_{n\in S_k}x_n\mathrm{e}^{-x_n/x_{n+1}}\ge \sum_{k=0}^\infty2^{-k}(|S_k|-1)\mathrm{e}^{-2}\ge \mathrm{e}^{-2}\sum_{k=0}^\infty 2^{-k}|S_k|-\mathrm{e}^{-2}\sum_{k=0}^\infty2^{-k}=\infty. $$

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  • $\begingroup$ This is a very nice answer! Thank you very much! $\endgroup$
    – ling
    Apr 4, 2023 at 11:40
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(This is inspired by Yiorgos S. Smyrlis' answer.)

Divide $\Bbb N$ into two complementary sets of indices $$ A = \{ n \in \Bbb N \mid x_{n+1} \le \frac 12 x_n \} \, , \\ B = \{ n \in \Bbb N \mid x_{n+1} > \frac 12 x_n \} \, . $$

If $k < l$ are two elements of $A$ then $x_l \le \frac 12 x_k$ (here we need the fact the the given sequence is decreasing). It follows that the $k$'th element of $A$ is $\le 2^{-k}x_1$. $A$ can be empty, finite, or infinite, but in any case is $$ \sum_{n \in A} x_n < \infty $$ and therefore $$ \sum_{n \in B} x_n = \infty \, . $$

Then $$ \sum_{n=1}^\infty x_n e^{-\frac{x_{n}}{x_{n+1}}} \ge \sum_{n \in B} x_n e^{-\frac{x_{n}}{x_{n+1}}} \ge e^{-2} \sum_{n \in B} x_n = \infty \, . $$


Remark: The proof shows that under the given conditions, $$ \sum_{n=1}^\infty x_n f\left(\frac{x_{n}}{x_{n+1}}\right)=\infty $$ holds for any positive, monotonic decreasing function $f:(0, \infty) \to (0, \infty)$.

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  • $\begingroup$ (+1) This is the way I approached this. It is simple and extends well. $\endgroup$
    – robjohn
    Sep 7, 2023 at 19:35
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    $\begingroup$ I had clicked, but it doesn't seem to have stuck. I tried again. $\endgroup$
    – robjohn
    Sep 8, 2023 at 9:52

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