# If $\{x_n\}$ is positive, decreasing and $\sum x_n=\infty,$ then $\sum x_ne^{-\frac{x_{n}}{x_{n+1}}}=\infty$

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.

• 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. Apr 4, 2023 at 8:49
• Could you give more details on what you tried? Apr 4, 2023 at 13:08

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 $$S_k=\{n\in\mathbb N: 2^{-k} Then $$\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.$$

• This is a very nice answer! Thank you very much!
– ling
Apr 4, 2023 at 11:40

(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 $$f:(0, \infty) \to (0, \infty)$$.

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