I did the following
Problem: Let $(a_n)$ be a convergent sequence with $\lim_{n \to\infty} a_n=a$ then the sequence $(b_n)$ defined by $b_n:=\frac{1}{n+1}(a_0+a_1+...+a_n)$ also converges to $a$.
Since I came up with a different solution to one given, I post it here and ask if this is correct.
Solution: We can rewrite the defintion of $b_n$ to $b_n=\sum_{k=0}^n\frac{a_k}{n+1}$. Now we show that for every $\epsilon>0$ there exists a $n\in\mathbb{N}$ with
\begin{align}|b_n-a|&=|\sum_{k=0}^n\frac{a_k}{n+1} -a|\\ &=|\frac{1}{n+1}\left( \sum_{k=0}^n a_k -\sum_{k=0}^n a\right) |\\ &\le\frac{1}{n+1}\sum_{k=0}^n | a_k - a |\\ &<\epsilon\end{align}
if $n\ge N$.
Now since $(a_n)$ is convergent and bounded, there exists a $M$ with
$$k\ge M \implies |a_k-a|<\frac{\epsilon}{2}$$ $$k < M \implies |a_k-a|\le M_k$$
It follows that
\begin{align}\frac{1}{n+1}\sum_{k=0}^n | a_k - a |\\ &=\frac{1}{n+1}\left(\sum_{k=0}^{M-1} | a_k - a | -\sum_{k=m}^{n} | a_k - a |\right)\\ &\le\frac{M}{n+1} M_k +\frac{(n-M)}{n+1}\frac{\epsilon}{2}\end{align}
Then, there exists a $N>M$ so that $n \ge N\implies \frac{M}{n+1} M_k<\frac{\epsilon}{2}$ and we have
$$\frac{1}{n+1}\sum_{k=0}^n | a_k - a |<\frac{\epsilon}{2}+\frac{\epsilon}{2}<\epsilon$$
which shows, that $(b_n)$ is convergent. $\blacksquare$