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.

  • 1
    $\begingroup$ 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

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 $\sum_{k=0}^\infty b_{9k} = \infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.