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Let the sequence $a_n=b_1(n)+b_2(n)$ I know that if the series $\displaystyle\sum_{n=1}^\infty b_1(n)$ is convergent and the series $\displaystyle\sum_{n=1}^\infty b_2(n)$ is convergent then the series $\displaystyle\sum_{n=1}^\infty a_n$ is convergent. Now we can show by induction that given $a_n=b_1(n)+b_2(n)+\cdots+b_N(n)$ with the series $\displaystyle\sum_{n=1}^\infty b_i(n)$ is convergent for all $i=1..N$ then the series $\displaystyle\sum_{n=1}^\infty a_n$ is convergent. My question is the following suppose that $a_n=b_1(n)+b_2(n)+b_3(n)+\cdots$, I mean $a_n$ is an infinite sum of $b_i(n)$ for all integer $i\ge 1$. Suppose that we know that the series $\displaystyle\sum_{n=1}^\infty b_i(n)$ is convergent for all integer $i\ge 1$, can we say that the series $\displaystyle\sum_{n=1}^\infty a_n$ is convergent ? I'm asking because I want to write the power series representation of $a_n$ (seen as a function of $n$), so that $a_n=b_1(n)+b_2(n)+\cdots$ and then arguing that all the series $\displaystyle\sum_{n=1}^\infty b_i(n)$ are convergent for all integer $i\ge 1$ and hence the series $\displaystyle\sum_{n=1}^\infty a_n$ is convergent and I want to do this without the big O notation. thank you for your help!

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  • $\begingroup$ What if you take something like $$b_k(n) = \begin{cases} \frac{1}{n^2} &\text { if } n\leq k\\ 0 & \text{ otherwise.} \end{cases}$$ ? $\endgroup$ – Clement C. Feb 15 '16 at 19:03
  • $\begingroup$ Dear Clement, Is this a counter example ? I don't get it $\endgroup$ – palio Feb 15 '16 at 20:28
  • $\begingroup$ Yes, see below. $\endgroup$ – Clement C. Feb 15 '16 at 22:40
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Below is a counter example.* Take $$b_k(n) = \begin{cases} 1 &\text { if } n\leq k\\ 0 & \text{ otherwise.} \end{cases}$$

Then $\sum_{n=1}^\infty b_k(n)=\sum_{n=1}^k b_k(n) = k$ is finite for all $k\geq 1$, but $$a_n = \sum_{k=1}^\infty b_k(n)=\sum_{k=n}^\infty b_k(n) = \sum_{k=n}^\infty 1 = \infty$$ so a fortiori $\sum_{n=1}^\infty a_n = \infty$.

${}^\ast$ Provided I did answer your question and did not mess up in the double-indexing.

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  • $\begingroup$ It does not answer my question since you took a divergent series from the start: the series with general term $1$ is divergent. I wanted all the series with general term $b_k(n)$ to be convergent. so taking $b_k(n)=1$ does not match the case in study. Actually we know that the sum of convergent series with a divergent series is always divergent series so I don't expect that the infinite sum of mixed convergent and divergent series to be convergent $\endgroup$ – palio Feb 15 '16 at 22:59
  • $\begingroup$ Read my answer... the series are convergent. @palio $\endgroup$ – Clement C. Feb 15 '16 at 23:25

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