Reed Eaton

2022-06-27

If ${y}_{n}>0$, $lim\frac{{x}_{n}}{{y}_{n}}=a$ and $\sum {y}_{n}=+\mathrm{\infty }$, then $lim\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{{y}_{1}+{y}_{2}+...+{y}_{n}}=a$

Trey Ross

Expert

You had a good idea. Indeed, since $\frac{{x}_{n}}{{y}_{n}}\to a$, then for any $\epsilon >0$ there is some N such that for $k\ge N$ you get $\left(a-\epsilon \right){y}_{k}\le {x}_{k}\le \left(a+\epsilon \right){y}_{k}$ (we've used that ${y}_{k}>0$ here, so that the signs of inequalities remain). Then for $n\ge N$ you can rewrite
$\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}=\frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...+{y}_{n}}+\frac{{x}_{N+1}+...{x}_{N}}{{y}_{1}+...+{y}_{n}}\le \frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...{y}_{n}}+\left(a+\epsilon \right)\frac{{y}_{N+1}+...+{y}_{n}}{{y}_{1}+...+{y}_{n}}$
Now, with N fixed, you have $\frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...+{y}_{n}}\to 0$ as $n\to \mathrm{\infty }$, due to divergence of series $\sum _{k}{y}_{k}$. Moreover,
$\frac{{y}_{N+1}+...{y}_{n}}{{y}_{1}+...+{y}_{n}}=1-\frac{{y}_{1}+...+{y}_{N}}{{y}_{1}+...+{y}_{n}}\underset{n\to \mathrm{\infty }}{\overset{}{\to }}1-0=1$
due to the same reason. Hence $\underset{n}{lim sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a+\epsilon$. Similarly you can prove that $\underset{n}{lim inf}\frac{{x}_{1}+...{x}_{n}}{{y}_{1}+...+{y}_{n}}\ge a-\epsilon$ and since $\epsilon >0$ was arbitrary you can conclude
$a\le \underset{n}{lim inf}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le \underset{n}{lim sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a$
and so the limit exists and is equal to a.

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