Prove that for any $k\in \mathbb{N}$, the series $\sum a_{n}^{k}b_{n}$ is convergent. Let $\left ( a_{n} \right )$ and $\left ( b_{n} \right )$ be two sequences of real numbers satisfying:

*

*The partial sums of $\left ( b_{n} \right )$ is bounded: there exists $L>0$ such that $\left | b_{1} +\cdot \cdot \cdot +b_{k}\right |<L$ for any $k$,

*$\lim a_{n}=0$

*$\sum \left | a_{n}-a_{n+1} \right |$ converges.

Prove that for any $k\in \mathbb{N}$, the series $\sum a_{n}^{k}b_{n}$ is convergent.
My approach:
Let $\varepsilon >0$.
Since $\lim a_{n}=0$, for $\left ( \frac{\varepsilon }{3L} \right )^{\frac{1}{k}}> 0$, $\exists N_{1}$ s.t $\left | a_{n} \right |<\left ( \frac{\varepsilon }{3L} \right )^{\frac{1}{k}} \; \forall n>N_{1}$. $\Rightarrow \left | a_{n} \right |^{k}<\frac{\varepsilon }{3L} \; \; \; \forall k$.
Let $B_{k}=b_{1}+\cdot \cdot \cdot +b_{k}$, then $\left | B_{k} \right |<L$ for any $k$.
Then, $\left | a_{n} ^{k}B_{{k}'}\right |=\left | a_{n} \right |^{k} \left | B_{{k}'} \right |<\frac{\varepsilon }{3}$ for any ${k}'\in \mathbb{N}$
Since $\sum \left | a_{n}-a_{n+1} \right |$ converges, $\lim \left | a_{n}-a_{n+1} \right |=0$.
Then, for arbitrary $m,l\in \mathbb{N}$ s.t $(m-l)\geq 1$, and for $\frac{3L}{\varepsilon (m-l)^{2}k}>0$, $\exists N_{2}$ s.t $\left | a_{n}-a_{n+1} \right |<\frac{3L}{\varepsilon (m-l)^{2}k}\; \; \; \forall n> N_{2}$.
Let $N=\max \left ( N_{1},N_{2} \right )$
Then, $\forall\: \:  m-1\geq l>N$, we have $$\left | \sum_{n=l}^{m-1}B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right ) \right |\leq \sum_{n=l}^{m-1}\left | B_{n} \right |\left | a_{n}-a_{n+1} \right |\left | a_{n}^{k-1}+a_{n}^{k-2}a_{n+1}+\cdot \cdot \cdot +a_{n}a_{n+1}^{k-2}+a_{n+1}^{k-1} \right |$$
$$\leq (m-l)L\cdot (m-l)\frac{3L}{\varepsilon (m-l)^{2}k}\left ( \left | a_{n}^{2k} \right |+\left |a_{n}^{k} a_{n+1}^{k} \right |+\cdot \cdot \cdot +\left | a_{n}^{k}a_{n+1}^{k} \right | +\left | a_{n+1}^{2k} \right |\right )$$ $$<(m-l)L\cdot (m-l)\frac{3L}{\varepsilon (m-l)^{2}k}\cdot k\frac{\varepsilon ^{2}}{9L^{2}}=\frac{\varepsilon }{3} $$
Also, $\forall m\geq l> N$,
$\sum_{n=l}^{m}a_{n}^{k}b_{n} = \sum_{n=l}^{m}a_{n}^{k}\left ( B_{n}-B_{n-1} \right )$ $$=a_{l}^{k}B_{l}-a_{l}^{k}B_{l-1}+a_{l+1}^{k}B_{l+1}-a_{l+1}^{k}B_{l}+a_{l+2}^{k}B_{l+2}-a_{l+2}^{k}B_{l+1}+\cdot \cdot \cdot +a_{m}^{k}B_{m}-a_{m}^{k}B_{m-1}$$
$$=-a_{l}^{k}B_{l-1}+a_{m}^{k}B_{m}+\sum_{n=l}^{m}B_{n}\left ( a_{n}^{k} -a_{n+1}^{k}\right )$$
$\Rightarrow \left | \sum_{n=l}^{m}a_{n}^{k}b_{n} \right |\leq \left | -a_{l}^{k}B_{l-1} \right |+\left | a_{m}^{k}B_{m} \right |+\left | \sum_{n=l}^{m-1}B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right ) \right |$ $<\frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}=\varepsilon $
$\Rightarrow \sum a_{n}^{k}b_{n}$ satisfies the cauchy criterion.
$\Leftrightarrow \sum a_{n}^{k}b_{n}$ converges. $\blacksquare $
Does my approach make sense?
I would be appreciated for any feedback on this proof, thank you.
P.S. I know that we can use the Triangle Inequality as I did in the proof for any absolute convergent series, but when I look at some experts' proofs, it seems like they are using it without considering if a series converges absolutely or not (maybe I'm wrong.) I just want to clarify if I could use the Triangle Inequality for this problem.
Edited: I tried to fix my solution with Michał Miśkiewicz's answer, but it is really hard to bound $\left | \sum_{n=l}^{m-1}B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right ) \right |$ $<\frac{\varepsilon }{3}$ like this.
I hope I can get some help with this.

*

*Here is what I tried to bound $\left | \sum_{n=l}^{m-1}B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right ) \right |$.

Since $\sum \left | a_{n}-a_{n+1} \right |$ converges, it satisfies the Cauchy Criterion.
So, for arbitrary $x$,$y$ s.t $x-y \geq 1$, and for each $\left ( \frac{\varepsilon }{3L\left ( x-y \right )} \right )^{\frac{1}{k}}>0$, $\exists N_{i}  $ $(i=2,3,4,...)$ s.t $\left | \sum_{n=x}^{y}\left | a_{n}-a_{n+1} \right | \right |= \sum_{x}^{y}\left | a_{n}-a_{n+1} \right |< \left ( \frac{\varepsilon }{3L(x-y)} \right )^{\frac{1}{k}}\: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | \sum_{x}^{y} \left ( a_{n}-a_{n+1} \right )\right |<\left ( \frac{\varepsilon }{3L(x-y)} \right )^{\frac{1}{k}}$ $\forall y\geq x>N_{i}$
$\Rightarrow \left | a_{x}-a_{x+1}+a_{x+1}-a_{x+2}+\cdot \cdot \cdot +a_{y}-a_{y+}  \right |= \left |a_{x}-a_{y+1}  \right |<\left ( \frac{\varepsilon }{3L(x-y)} \right )^{\frac{1}{k}}\: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | a_{x}-a_{y+1}  \right |^{k}< \frac{\varepsilon }{3L(x-y)} \: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | a_{x}^{k}+(-1)^{k}a_{y+1}^{k}+\cdot \cdot \cdot  \right |<\frac{\varepsilon }{3L(x-y)} \: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | a_{x}^{k} +(-1)^{k}a_{y+1}^{k}\right |<\frac{\varepsilon }{3L(x-y)} \: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | a_{x}^{k}-a_{y+1}^{k} \right |<\frac{\varepsilon }{3L(x-y)} \: \: \: \; \forall y\geq x>N_{i}$
$\Rightarrow \left | a_{n}^{k}-a_{n+1}^{k} \right |<\frac{\varepsilon }{3L(x-y)}\: \: \: \; \forall n> N_{i},\forall y\geq x>N_{i}$
$\Rightarrow \left | B_{n} \right |\left | a_{n}^{k}-a_{n+1}^{k} \right |< \frac{\varepsilon }{3(x-y)}\: \: \: \; \forall n> N_{i},\forall y\geq x>N_{i}$
Let $N=\max \left ( N_{1},N_{2},N_{3},\cdot \cdot \cdot  \right )$, then $\forall m-1\geq l>N$ we have $$\left | \sum_{n=l}^{m-1} B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right )\right |\leq \sum_{l}^{m-1}\left | B_{n} \right |\left | a_{n}^{k}-a_{n+1}^{k} \right |<\frac{\varepsilon }{3(m-1-l)}(m-l)<\frac{\varepsilon }{3(m-l)}(m-l)=\frac{\varepsilon }{3}$$
Since we have $\left | \sum_{n=l}^{m-1} B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right )\right |<\frac{\varepsilon }{3}$, we can finish the proof.
Does my fixed version make sense? I feel like I'm making the same mistake as I did before.
 A: $(a_n)$ is convergent and therefore bounded, say $|a_n| < M$ for all $n$. Then
$$
 a_n^k-a_{n+1}^k = (a_n-a_{n+1})(a_n^{k-1} + a_n^{k-1}a_{n+1} + \cdots + a_{n+1}^{k-1})
$$
implies
$$
 |a_n^k-a_{n+1}^k| \le k M^{k-1} |a_n-a_{n+1}|,
$$
for $n \ge N$, and therefore
$$
\left | \sum_{n=l}^{m-1} B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right )\right |
\le  \sum_{n=l}^{m-1} |B_{n}|\left | a_{n}-a_{n+1} \right  |
\le L kM^{k-1} \sum_{n=l}^{m-1} \left| a_{n}-a_{n+1} \right  | \, .
$$
But $\sum \left | a_{n}-a_{n+1} \right |$ is convergent so that
$$
\sum_{n=l}^{m-1} \left| a_{n}-a_{n+1} \right  | < \frac{\epsilon}{3L kM^{k-1}}
$$
for $m > l \ge N$ with sufficiently large $N$. It follows that
$$
\left | \sum_{n=l}^{m-1} B_{n}\left ( a_{n}^{k}-a_{n+1}^{k} \right )\right |
< \frac{\epsilon}{3}
$$
for $m > l \ge N$. This shows (together with your other estimates) that $\sum a_{n}^{k}b_{n}$ satisfies the Cauchy criterion and is therefore convergent.

Note that you could also prove the theorem for $k=1$ only, and then show that if $(a_n)$ satisfies the given conditions then $(a_n^k)$ satisfies the same conditions.
A: The approach itself is adequate, but your reasoning contains a conceptual mistake. To see this, it's enough to notice you haven't really used the assumptions. What you actually used in the reasoning is:

*

*sequence $B_n$ is bounded,

*sequence $a_n$ tends to zero,

*sequence $|a_n-a_{n-1}|$ tends to zero.

Note that $|a_n-a_{n-1}| \to 0$ follows from $a_n \to 0$, so it's not really useful. Without exploiting the assumption $\sum \left | a_{n}-a_{n+1} \right | < \infty$ one cannot expect to prove convergence of $\sum a_n b_n$ (try and find a counterexample!), so there must be a mistake.
And indeed, there is one. For fixed $m,l$ you choose large enough $N_2$, and then pick $N$ as $N = \max(N_1,N_2)$. But after that, you are assuming that $m-1 > l > N$, as if $m$ and $l$ weren't already fixed.

But as I said, the approach makes sense - one can easily use summation by parts in the way you described, along with the estimate
$$|a_n^k-a_{n+1}^k| \le k \max(|a_n|,|a_{n+1}|)^{k-1} |a_n-a_{n+1}|,$$
to complete the proof. I suggest looking at the proof of Dirichlet's criterion. Note how this criterion is a special case of your problem, for $k=1$ and $a_n$ monotone.

Edit. There appeared a new version of the reasoning, so let me comment on this too. The mistake from before is still there, only $m$ and $l$ changed to $x$ and $y$. You fix $x,y$, which fixes the value of $\left ( \frac{\varepsilon }{3L\left ( x-y \right )} \right )^{\frac{1}{k}}$, and then pick sufficiently large $N$ according to this value. But then, you say some inequality holds for all $y > x \ge N$, which means you're using the same symbols for very different things, again.
Also, you consider a sequence $N_1,N_2,\ldots$ (indexed by $i$), but I couldn't find this $i$ elsewhere, so I see no point in introducing it.
For a correct solution to the problem, I recommend the other answer.
