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Suppose $ \lim_m \sum_n f(n,m) = c $ and $ 0 \leq c< \infty $. Is it true that $ \lim_m \sum_n f(n,m)^k =0 $ if k >1?

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2 Answers 2

up vote 3 down vote accepted

There are very simple counterexamples. For instance, if $f(n,m)=2^{-n}$ for all $m,n\in\Bbb N^2$, then

$$\lim_{m\to\infty}\sum_{n\ge 0}f(n,m)=\lim_{m\to\infty}\sum_{n\ge 0}\frac1{2^n}=\lim_{m\to\infty}2=2\;,$$

and

$$\lim_{m\to\infty}\sum_{n\ge 0}f(n,m)^2=\lim_{m\to\infty}\sum_{n\ge 0}\frac1{4^n}=\lim_{m\to\infty}\frac43=\frac43\;,$$

not $0$.

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Thanks for your answer –  user55449 Apr 11 '13 at 9:07
    
@user55449: You’re welcome. –  Brian M. Scott Apr 11 '13 at 9:08

The answer is obviously no, take for example:

$f(n,m)=1 $ if $n=1$ and $f(n,m)=0$ if $n\geq2$ then we have $\sum_n f(n,m)=1$ and forall $k>1$ $\sum_n f(n,m)^k=1\neq0$

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Thanks for your answer –  user55449 Apr 11 '13 at 9:08
    
You're welcome. –  Sami Ben Romdhane Apr 11 '13 at 9:47

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