# $a_{n}$ converges and $\frac{a_{n}}{n+1}$ too?

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 math.stackexchange.com/questions/207910/… 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.) – Jonas Meyer Dec 28 '12 at 8:18
thanks Jonas, great! – doniyor Dec 28 '12 at 8:21

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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yeah, but $b_n$ should also converge to $a$, this is what i need to show. how is that possible here? – doniyor Dec 28 '12 at 8:01
It does not converge to $a$. Perhaps you have copied the problem incorrectly, or your teacher made an error. – Alex Becker Dec 28 '12 at 8:02
oh okay, but the original problem is this: $\frac{1}{n+1}(a_0+a_1+a_2+...+a_n)$ But does it doesnot differ from $\frac{1}{n+1}(a_n)$ in terms of limit, right? – doniyor Dec 28 '12 at 8:06
@doniyor: Not right. That sequence does converge to $a$. Perhaps you should ask this as a new question since this one has already been answered. – Jonas Meyer Dec 28 '12 at 8:07
@doniyor The number of terms goes to $\infty$, not the sequence itself. Briefly, the reasoning is that as $n$ gets large, almost all of the terms $a_i$ for $i=0,1,\ldots,n$ are roughly $a$, so the sum $a_0+\cdots+a_{n+1}$ is roughly $(n+1)a$. – Alex Becker Dec 28 '12 at 8:17

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, $\lim \frac{a_n}{n+1} = (\lim a_n) (\lim \frac{1}{n+1})=0$.

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