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Let $(b_n)$ be a sequence where $\forall n \in \mathbb N \ 0 < b_{n+1} < b_n$ and $b_n \rightarrow 0$ as $n \rightarrow \infty$. Furthermore suppose that $\sum \limits_{n=1}^\infty b_n = \infty$.

What can I say about $\sum \limits_{n=9k, \ k \in \mathbb N}^\infty b_n$?

I'll refer to $B_m = \sum \limits_{n=1}^m b_n$ and $S_m = \sum \limits_{n=9k, \ k \in \mathbb N}^m b_n$.

I know that a monotonic and divergent sequence has no convergent subsequence. But my problem here is that $S_m$ is not a subsequence of $B_m$ so I don't know what tools to apply.

Examples I've been considering are the harmonic series and $b_n = \log{\frac{n}{n-1}}$.

This isn't homework, it's just related to something that I've been wondering about. Thanks for any help.

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    $\begingroup$ The sub-sum diverges. Since your sequence is decreasing and positive, you can bound your original sum from above and below by suitable multiples of your sub-sum (plus some additive constant). So the sub-sum and the original sum either converges at the same time or diverges at the same time. $\endgroup$ – achille hui Sep 16 '15 at 14:53
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Let $b_0 := b_1 +1$ (this makes notation a little more easy, but does not change the argument). Note that, as $(b_n)$ is decreasing, that \begin{align*} \sum_{n=0}^\infty b_n &= \sum_{k=0}^\infty \sum_{i=0}^8 b_{9k+i} \\ &\le \sum_{k=0}^\infty \sum_{i=0}^8 b_{9k}\\ &= 9\sum_{k=0}^\infty b_{9k} \end{align*} Hence, as $\sum_{n=0}^\infty b_n = \infty$, we have $\sum_{k=0}^\infty b_{9k} = \infty$.

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