Reed Eaton

Answered

2022-06-27

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

Answer & Explanation

Trey Ross

Expert

2022-06-28Added 30 answers

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 $(a-\epsilon ){y}_{k}\le {x}_{k}\le (a+\epsilon ){y}_{k}$ (we've used that ${y}_{k}>0$ here, so that the signs of inequalities remain). Then for $n\ge N$ you can rewrite

$\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}=\frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...+{y}_{n}}+\frac{{x}_{N+1}+...{x}_{N}}{{y}_{1}+...+{y}_{n}}\le \frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...{y}_{n}}+(a+\epsilon )\frac{{y}_{N+1}+...+{y}_{n}}{{y}_{1}+...+{y}_{n}}$

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\u2006sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a+\epsilon $. Similarly you can prove that $\underset{n}{lim\u2006inf}\frac{{x}_{1}+...{x}_{n}}{{y}_{1}+...+{y}_{n}}\ge a-\epsilon $ and since $\epsilon >0$ was arbitrary you can conclude

$a\le \underset{n}{lim\u2006inf}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le \underset{n}{lim\u2006sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a$

and so the limit exists and is equal to a.

$\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}=\frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...+{y}_{n}}+\frac{{x}_{N+1}+...{x}_{N}}{{y}_{1}+...+{y}_{n}}\le \frac{{x}_{1}+...+{x}_{N}}{{y}_{1}+...{y}_{n}}+(a+\epsilon )\frac{{y}_{N+1}+...+{y}_{n}}{{y}_{1}+...+{y}_{n}}$

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\u2006sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a+\epsilon $. Similarly you can prove that $\underset{n}{lim\u2006inf}\frac{{x}_{1}+...{x}_{n}}{{y}_{1}+...+{y}_{n}}\ge a-\epsilon $ and since $\epsilon >0$ was arbitrary you can conclude

$a\le \underset{n}{lim\u2006inf}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le \underset{n}{lim\u2006sup}\frac{{x}_{1}+...+{x}_{n}}{{y}_{1}+...+{y}_{n}}\le a$

and so the limit exists and is equal to a.

Most Popular Questions