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let

$ \sum_{n=1}^{\infty}a_{n}=1+\frac{1}{2}-2\cdot\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-2\cdot\frac{1}{6}+\frac{1}{7}+\frac{1}{8}-2\cdot\frac{1}{9}+... $

Meaning, $ \sum_{n=1}^{\infty}a_{n} $ is the seires which we multiplyed each third element of the harmonic series by $ -2 $.

I have to tell if this series converge/diverge.

Here's what I tried:

First, I defined $ S_{p}=\sum_{n=1}^{p}a_{n} $

and then I defined $ b_{k}=\frac{1}{3k-2}+\frac{1}{3k-1}-\frac{2}{3k} $ ( I summed each 3 elements of the original series and named it $ b_k $ ).

Thus, $ T_{P}=\sum_{k=1}^{p}b_{k}=\sum_{n=1}^{3p-2}a_{n}=S_{3p-2} $

So basiclly $ T_p $ is a subsequence of $ S_p $, therefore its enough to show that $ \sum_{k=1}^{\infty}b_{k} $ diverges, and it will imply that also $ \sum_{n=1}^{\infty}a_{n} $ diverges, because otherwise $ S_p $ is a convergent sequence and eny subsequence of $ S_p $ should converge to the same limit.

Thats my idea, it required a bit of work and eventually I managed to show that $ \sum_{k=1}^{\infty}b_{k}=\sum_{k=1}^{\infty}\frac{9k^{3}+9k^{2}-4k}{3k\left(3k-1\right)\left(3k-2\right)} $ diverges.

I wander if there's simplest way to prove that this series converges (if it is a convergent series at all).

Thanks in advance.

FreeZe
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  • $b_k \xrightarrow{k\to \infty}{\frac{1}{3}}$, so $\sum_k b_k$ diverges. – M-S Aug 14 '20 at 10:30
  • @metamorphy can you prove it? – FreeZe Aug 14 '20 at 13:59
  • You basically did it already. An easier way to see $b_k=\mathcal{O}(k^{-2})$ (which is sufficient for the convergence of $\sum_{k=1}^\infty b_k$ and hence of the original series) is directly from $$b_k=\left(\frac{1}{3k-2}-\frac{1}{3k}\right)+\left(\frac{1}{3k-1}-\frac{1}{3k}\right)=\frac{2}{3k(3k-2)}+\frac{1}{3k(3k-1)}.$$ – metamorphy Aug 14 '20 at 14:08
  • I see where I was wrong and why the series of $ b_k $ converge. But how is this imply that the original series also converge? is this because $ \sum_{k=1}^{p}b_{k}=S_{3p-2} $ where $ S_{p}=\sum_{n=1}^{p}a_{n} $, and since $ a_{n}\underset{n\to\infty}{\longrightarrow}0 $ we can use limit arithmethic to show it? – FreeZe Aug 14 '20 at 14:19

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