I am trying to show that if a sequence eventually increases less than another sequence, then if the infinite sum of the second sequence converges, so must the infinite sum of the first sequence. More formally:
Let $x_n >0 $ and $y_n >0$ for all $n∈\Bbb N$. Suppose that for some $N∈\Bbb N$ we have $\frac{x_{n+1}}{x_n} \le \frac{y_{n+1}}{y_n}$ for all $n \ge N$. Show that if $\sum y_n$ converges, then so does $\sum x_n$.
I have made some progress. Clearly $(y_n) \to 0$ and I've used this to show that $(x_n) \to 0$, though I know that's insufficient. I've been told that the best way is to find a way to bound $x_n$, but so far I've only been able to do that in terms of $x_{n+1}$ by noting that $\frac{y_{n+1}}{y_n}$ is bounded because it converges, which isn't very helpful.
Interestingly, this seems to be a reasonably useful result: comparing ratios of series to show series convergence, but I haven't found it referenced anywhere, so is this perhaps a trivial special case of some more general test?
Thank you
Note that you can rearrange this inequality as $$ x_{n+1}\leq \frac{y_{n+1}}{y_n}x_n\leq\frac{y_{n+1}}{y_n}\frac{y_n}{y_{n-1}}x_{n-1}\leq\frac{y_{n+1}}{y_n}\frac{y_n}{y_{n-1}}\frac{y_{n-1}}{y_{n-2}}x_{n-2}\leq\cdots. $$ This tells you, in particular, that for any fixed $m<n$, for any $k\in[m,n]$ we have $$ x_k\leq\frac{y_k}{y_m}x_m, $$ so that $$ \begin{align*} \sum_{k=m}^{n}x_k\leq\sum_{k=m}^{n}\frac{y_k}{y_m}x_m=\frac{x_m}{y_m}\sum_{k=m}^{n}y_k. \end{align*} $$ Because the terms are all positive, you can use this inequality to show that whenever $\sum y_k$ is Cauchy, so is $\sum x_k$.