$a_{n}$ 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? 

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