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So, here we go again, the sequence $x_n$ is increasing and $x_n\to\infty$ as $n\to\infty$, and also, $\lim\limits_{n\to\infty}\dfrac{x_{n+1}}{x_n}= x$ which is a real non zero number,

Prove that :

$$\lim\limits_{n\to\infty}\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} = x $$

I'm stuck again, I know why it says that [eventually] $x_n>A$ for every $A$, from that I got :

$$\frac{x_1+\cdots+x_{n+1}}{x_1+\cdots+x_n} -x\leq (n+1)\frac{x_{n+1}}{x_n} -x$$ for every $n>N$

$N$ is special though but I don't get anywhere from there so it doesn't matter. can anybody help?

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  • $\begingroup$ doesnt anybody have any ideas? im really stuck with this one $\endgroup$ – Plom Jul 24 '13 at 1:50
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Stolz-Cesaro says

Suppose $b_n$ is a sequence of strictly increasing numbers such that $b_n\nearrow +\infty$ and let $a_n$ be any sequence of real numbers. If $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\to\ell $$ then $$\frac{a_n}{b_n}\to\ell$$

Now, let $$a_n=\sum_{k=1}^{n} x_k$$ $$b_n=\sum_{k=1}^{n-1} x_k$$

Then $$a_{n+1}-a_n=x_{n+1}\\ b_{n+1}-b_n=x_{n}$$

and $b_n\nearrow +\infty$, $b_n$ is strictly increasing.

For a proof, see here.

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