Can anyone help me solve this problem on Sequences/Series?

I would be so grateful if someone helped me solve these 2 questions about series and sequences: $$\text{Let } (u_n)_{n\in\mathbb{N}} \text{ a real sequence, and } (v_n)_{n\in\mathbb{N}} \text{ the sequence defined as :}$$ $$v_n = \frac{1}{n+1} \sum_{k=0}^n u_k$$

1. Prove that if $(v_n)_{n\in\mathbb{N}}$ is bounded, $(u_n)_{n\in\mathbb{N}}$ isn't necessarily bounded.
2. Prove that if $(u_n)_{n\in\mathbb{N}}$ is increasing, then $(v_n)_{n\in\mathbb{N}}$ is too.

For the first one, I have not really gotten that far, because I didn't even know how to approach it. And for the second one, I tried to simplify the difference $\frac{1}{n+2}\sum_{k=0}^{n+1} u_k - \frac{1}{n+1}\sum_{k=0}^n u_k$, but I didn't get anywhere.

I have also started from $u_k \leq u_{k+1}$ getting to $\sum_{k=0}^{n} u_k \leq \sum_{k=0}^{n+1} u_k$, but I don't really know what to do after that.

Any help would be greatly appreciated even if just a clue.

Thanks!

For Part 1, to get $v_n$ bounded but $u_n$ unbounded, the idea is to have $|u_n|\le 1$ for "almost all" $n$ and to have $|u_n|$ "large" but not too large, for a "sparse" set of $n.$

E.g. suppose $u_n=0$ when $n$ is not a power of $4,$ but if $0\le m\in\Bbb Z$ and $n=4^m$ then $u_n=\sqrt n.$ Now for any $n> 0,$ take $m\in \Bbb Z$ such that $$4^{m-1}\le n<4^m.$$ Then we have $$0\le \sum_{k=0}^nu_k<\sum_{k=0}^{4^m}u_k=$$ $$=\sum_{j=0}^m2^j=2^{m+1}-1<2^{m+1}=2(2^m)\le 2\sqrt {4n}$$ because $4^{m-1}\le n\implies 2^m=\sqrt {4^m}\le \sqrt {4n}.$

So $v_n$ actually converges to $0,$ but there are infinitely many $n$ for which $u_n=\sqrt n.$

If $u_n$ is bounded, then there exists $M$ s.t. $|u_n|<M$ for all $n$. Thus $$ v_n \leq \frac{(n+1)M}{n+1} = M$$ So $v_n$ is bounded.

As for the second part, as with all maths problems, begin with your intuition. It's saying that (essentially) the mean of an increasing sequence, adding terms one by one, is itself increasing. Think: Why is this true? There will be many ways to solve this but the way I went to first was realising that if $u_n$ is increasing, then not only is it larger than the previous term, but also larger than all the previous terms. I'd then use your first method to prove the result.

EDIT: so now the first part is to find an unbounded $u_n$ such that their mean converges. You asked for clues so I won't go into detail but I'd have thought this could work with a sequence that's mostly constant but with the occasional increasingly large number. The reason I thought of this is that a constant function is intuitively the only one that'll grow slowly enough,$^\dagger$ but obviously that is bounded. So we need to have a "mostly constant" function.

As a further hint to your second part, note that you can write $$u_{n+1} \geq \frac{1}{n+1} \sum_{k=0}^n u_{n}$$ This should help you expand your difference of sums.

$\dagger$ – because the derivative of $x$ is a constant.