-1
$\begingroup$

how to derive that $\sum a_j$ converges.

$\endgroup$

closed as off-topic by Saad, RRL, KReiser, DRF, Gibbs Dec 10 '18 at 10:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, RRL, KReiser, DRF, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Take a sum of each of the partial sums you have, starting from $n=1$. The range of those partial sums grows exponentially fast, and you will find that the appropriate terms on the RHS shrink fast enough to converge. $\endgroup$ – GenericMathematician Dec 9 '18 at 23:35
  • $\begingroup$ The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy. $\endgroup$ – Sean Roberson Dec 10 '18 at 0:00
1
$\begingroup$

We have that

$$\sum_{j=1}^{\infty} a_j=\sum_{k=0}^{\infty}\left(\sum_{j=2^k}^{2^{k+1}-1} a_j\right)\le \sum_{k=0}^{\infty}\left(\sum_{j=2^k}^{2^{k+1}} a_j\right)\le \sum_{k=0}^{\infty}\frac{1}{\sqrt{2^k}}$$

$\endgroup$
  • $\begingroup$ How does the first equality derive? $\endgroup$ – Tsubaki Dec 10 '18 at 0:00
  • $\begingroup$ @Tsubaki Let try with $k=0,1,2...$ and see what happens. $\endgroup$ – gimusi Dec 10 '18 at 0:01
  • $\begingroup$ Is it if k =1, its result equals to j = 3, and so on? but if so, how do you achieve to the next inequality? $\endgroup$ – Tsubaki Dec 10 '18 at 0:11
  • $\begingroup$ @Tsubaki for $k=0 \implies j=1$ for $k=1 \implies j=2 to 3$ for $k=2 \implies j=4 to 7$ and so on.Can you see the pattern? $\endgroup$ – gimusi Dec 10 '18 at 0:16
  • $\begingroup$ May I ask how latter part converges? $\endgroup$ – Tsubaki Dec 10 '18 at 0:55
0
$\begingroup$

Let $$b_n = \sum_{j = 2^{n-1}}^{2^n-1} a_j \leq\frac{1}{\sqrt{2^n}}$$ Then $\sum b_n = \sum a_n$ and $\sum b_n$ converges as $\sum \frac{1}{\sqrt{2^n}}$ converges.

$\endgroup$
  • $\begingroup$ May I ask how latter part converges? $\endgroup$ – Tsubaki Dec 10 '18 at 0:55

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