# How this series a_j converges? (duplicate, solve) [closed]

how to derive that $$\sum a_j$$ converges.

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

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• 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. – GenericMathematician Dec 9 '18 at 23:35
• The tail is bounded by something that can be made small. It should be enough to show the tail remainder is Cauchy. – Sean Roberson Dec 10 '18 at 0:00

We have that

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

• How does the first equality derive? – Tsubaki Dec 10 '18 at 0:00
• @Tsubaki Let try with $k=0,1,2...$ and see what happens. – gimusi Dec 10 '18 at 0:01
• 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? – Tsubaki Dec 10 '18 at 0:11
• @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? – gimusi Dec 10 '18 at 0:16
• May I ask how latter part converges? – Tsubaki Dec 10 '18 at 0:55

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.

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