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I have a sequence $a_{n}$ which converges to $a$, then I have another sequence which is based on $a_{n}$: $b_{n}:=\frac{a_{n}}{n+1}$, now I have to show that $b_{n}$ also converges to $a$.

My steps:

$\frac{a_{n}}{n+1}=\frac{1}{n+1}\cdot a_{n}=0\cdot a=0$ But this is wrong, why am I getting this? My steps seem to be okay according to what I have learned till now; can someone show me the right way please?

And then I am asked to find another two sequences like $a_n$ and $b_n$ but where $a_n$ diverges and $b_n$ converges based on $a_n$. I said: let $a_n$ be a diverging sequence then
$b_n:=\frac{1}{a_n}$ the reciprocal of $a_n$ should converge. Am I right?

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    See http://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means for a related question (which might have been intended here). (This was also linked in a comment below, but I wanted to make it more visible.)2012-12-28

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You are correct, although your proof is not. Since $a_n\to a$ and $\frac{1}{n+1}\to 0$, we have that $\lim_{n\to\infty} \frac{1}{n+1}a_n=\lim_{n\to\infty}\frac{1}{n+1}\cdot\lim_{n\to\infty}a_n=0\cdot a=0.$

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    @JonasMeyer, yeah, sorry i misread your statement. thanks$a$lot. Alex, great now i got it. i had problems with understanding the sum of sequences at once.. nice, man :)2012-12-28
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For $a_n=1$, clearly $a_n \to 1$ and $b_n \to 0$. So the result you are trying to prove is false.

In fact, because product is continuous, $\lim \frac{a_n}{n+1} = (\lim a_n) (\lim \frac{1}{n+1})=0$.