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Given $a_n, b_n$ such that $\exists N \forall n>N a_n, b_n \geq 0$ I want to show that if $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge then $\sum_{n=0}^{\infty}a_n b_n$ also converges.

I tried to show it in the following way, if $\sum_{n=0}^{\infty}b_n$ converges, then $\lim_{n\rightarrow\infty}b_n=0$, so there exists $n_0$ such that $b_{n_0}\geq b_n$ for all $n\gt n_0$, therefore $a_n b_{n_0}\geq a_n b_n$, so that $\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n$, and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning?

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    $\begingroup$ Your proof is fine. I misread your proof a bit the first time so the "simplification" I meantioned in a previous comment is not really a big simplification it's just the special case $1 \geq b_n$ instead of $b_{n0} \geq b_n$. This is just nitpick, what you have done is perfectly fine. If you want to be really precise you might mention the monotone convergence theorem which is what you use to deduce that it's convergent (increasing + bounded = convergent) $\endgroup$ – Winther Dec 29 '16 at 2:52
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$\sum_m^na_kb_k = b_n\sum_m^na_k + \sum_{k=m}^{n-1} \big(\sum_{j=m}^{k}a_j \big)(b_k - b_{k+1})$.

This is the abels formula. You can now use the bound on b and convergence of sequence $a_n$ to get the convergence.

Since $b_n$ sequence is convergent you can easily show a bound M on $b_n$ and then use Cauchy criterion to get an N for any given $\epsilon$ such that $\sum_m^na_k < \epsilon / 4M \hspace{2mm} \forall m,n > N$ which will give you

$|\sum_{k=m}^{n}a_kb_k| \leq M \frac{\epsilon}{4M} + 2M \frac{\epsilon}{4M} < \epsilon$

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Take $N'\geq N$ such that $n>N'\implies a_n\leq 1.$ This is possible because the convergence of $\sum_n a_n$ implies that $a_n\to 0$ as $n\to \infty.$ For all $n>N'$ we have $0\leq |a_nb_n|=a_nb_n\leq b_n=|b_n|.$

The absolute values of the terms of the series $\sum_{n>N'}a_nb_n$ do not exceed the absolute values of the terms of the absolutely convergent series $\sum_{n>N'}b_n,$ so $\sum_{n>N'}a_nb_n $ converges.

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There's another proof, involving partial sums:

By hypothesis you know that, $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge, then by definition, if $(S_n)_{n\in\mathbb{N}}$ and $(T_n)_{n\in\mathbb{N}}$ are respectively the partial sums then they converge. So, by a previous result, it is known that $\lim_{n \to \infty}(S_n \cdot T_n)$ exists, ie, $(S_n \cdot T_n)_{n \in \mathbb{N}}$ converges and that implies that $\sum_{n=0}^{\infty}a_nb_n$ converges.

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