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Prove that the series $\sum_{n=1}^{\infty} a_n b_n$ converges if the following conditions are met:

  1. series $\sum_{n=1}^{\infty} b_n$ converges,
  2. series $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ absolutely converges.

I was thinking applying Abel's test, proving that if:

  1. $\sum_{n=1}^{\infty} b_n$ is a convergent series (given),
  2. {$a_n$} is a monotone sequence, and
  3. {$a_n$} is bounded.

Then the $\sum_{n=1}^{\infty} a_n b_n$ converges.

To prove the second one statement, we've to get following inequality: $a_{n+1} \leq a_n$. I've no idea how do we do that.

Third one, I think, is obtained by the fact, that the $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ series converges, so $\lim_{n\to\infty} (a_n - a_{n+1}) = 0$. Is it correct?

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    $\begingroup$ Instead of using Abel's theorem itself, use its proof, or rather summation by parts. For a partial sum, beside the two terms that are left over when using summation by part and that tend to zero, since $\sum_{1}^{m} b_n\to B$ and $a_n-a_{n+1}\to0$, you also get a sum $\sum_{r}^{s} (a_{n}-a_{n+1})\sum_{1}^n b_k$. The sum you bound using triangle inequality by $\sum_{r}^{s} |a_n-a_{n+1}||\sum_{1}^{n}b_k|$ The factor $|\sum_{1}^{n}b_k|$ is bounded by some $M$. So you get the bound $M\sum_{r}^{s}|a_{n}-a_{n+1}|$ which can be small. $\endgroup$
    – plop
    Jun 16 at 14:49
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There is no way of proving that $(a_n)_{n\in\Bbb N}$ is monotonic, since it may well not be.

It's not hard to prove that you always have, when $m,n\in\Bbb N$ and $m\geqslant n$,\begin{multline}\sum_{k=n}^na_kb_k=(a_n-a_{n+1})b_n+(a_{n+1}-a_{n+2})(b_n+b_{n+1})+\cdots+\\+(a_m-a_{m+1})\left(\sum_{k=n}^mb_k\right)+a_{m+1}\left(\sum_{k=n}^mb_k\right).\end{multline}Note that the sequence $(a_n)_{n\in\Bbb N}$ converges. In fact\begin{align}\sum_{n=1}^\infty|a_n-a_{n+1}|\text{ converges}&\implies\sum_{n=1}^\infty(a_n-a_{n+1})\text{ converges}\\&\iff(a_n)_{n\in\Bbb N}\text{ converges.}\end{align}Since it converges, it is a bounded sequence and therefore the sequence $\bigl(|a_n|\bigr)_{n\in\Bbb N}$ is bounded too.

Now, let $\varepsilon>0$. I will prove that there is a natural number $N$ such that if $m\geqslant n\geqslant N$, then$$\left|\sum_{k=n}^ma_kb_k\right|<\varepsilon,\tag1$$in order to apply the Cauchy's convergence test. Take $\varepsilon'\in(0,\varepsilon)$ and take $N\in\Bbb N$ such that$$m\geqslant n\geqslant N\implies\left|\sum_{k=n}^mb_k\right|<\frac{\varepsilon'}{\displaystyle\sum_{k=1}^\infty|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|}.$$Actually, this doesn't make sense if every $a_n$ is $0$, but then the statement is trivially true. Now we have, thanks to the first equality of this proof, that, if $m\geqslant n\geqslant N$,\begin{align}\left|\sum_{k=n}^na_kb_k\right|&\leqslant\frac{\varepsilon'}{\displaystyle\sum_{k=1}^\infty|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|}\left(\sum_{k=n}^m|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|\right)\\&\leqslant\varepsilon'\\&<\varepsilon.\end{align}

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