2

Show that

$$\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n} = 3$$

I think I can do the following which I am not quite sure about:

$$\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n} =\sum_{n=1}^{\infty}\dfrac{1}{2^{n-\sqrt{n}}}+\dfrac{1}{2^{n+\sqrt{n}}}$$

But after this I am quite not sure how to form the two infinite GP series. Please help.

I have also learned that $$\sum_{n=1}^{[x]} a^{\sqrt{n}} = \dfrac{2}{(\log_e{a})^2}+c$$ for $a>0$ and $\in \mathbb{N}$

Does this help my problem anyhow? Snapshot of the original problem:

enter image description here

Saradamani
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    What does "the largest integer nearest to $\sqrt{n}$ mean? What is $\langle10 \rangle$? $3$ or $4$? – Robert Z Apr 19 '18 at 08:28
  • @RobertZ it has not been mentioned there. But let us consider 3. – Saradamani Apr 19 '18 at 08:32
  • You could use $\lfloor\sqrt n\rfloor$. –  Apr 19 '18 at 08:48
  • @YvesDaoust With $\langle n \rangle=\lfloor\sqrt n\rfloor$ the sum is not $3$. I suspect that here $\langle n \rangle=\text{round}(\sqrt{n})$. – Robert Z Apr 19 '18 at 09:19
  • @RobertZ: I agree. The OP wrongly phrased. –  Apr 19 '18 at 09:33
  • No @Yves Daoust the question is not wrongly phrased. It is exactly the same written in the book Test Of Mathematics at the 10+2 Level by Indian Statistical Institute,Kolkata. I can put a snapshot of the same as well if you do not believe my words. – Saradamani Apr 19 '18 at 09:36
  • No @Robert Z the question is not wrongly phrased. It is exactly the same written in the book Test Of Mathematics at the 10+2 Level by Indian Statistical Institute,Kolkata. I can put a snapshot of the same as well if you do not believe my words. – Saradamani Apr 19 '18 at 09:36
  • @Saradamani: please do. –  Apr 19 '18 at 09:42
  • No I never do that. My own OPs are downvoted. Why on earth I would do that? – Saradamani Apr 19 '18 at 09:46
  • There are some "morally upright" users who do that. They closed one question of mine which I myself had to answer. They put it on hold first and then close it- very bad users they are! – Saradamani Apr 19 '18 at 09:47
  • Anyway, the text says "the integer nearest to" Not the "largest integer". So it is the one I used in my answer. – Robert Z Apr 19 '18 at 09:47
  • Please edit your question with the correct definition of $\langle n \rangle$ – Robert Z Apr 19 '18 at 09:50
  • Rather than complaining about users, you should try and understand what they mean and in what way they are helping. –  Apr 19 '18 at 10:29

1 Answers1

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Hint. If $\langle n \rangle=\text{round}(\sqrt{n})$ ("round" is also called the nearest integer function) then $$\begin{align}\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}&=\sum_{k=1}^{\infty}\sum_{j=1}^{2k}\frac{2^k+2^{-k}}{2^{k^2-k+j}} =\sum_{k=1}^{\infty}\frac{2^k+2^{-k}}{2^{k^2-k}}(1-2^{-2k}) \\ &=2\sum_{k=1}^{\infty}(2^{-(k-1)^2}-2^{-(k+1)^2}).\end{align}$$ Now, in order to obtain the final result $3$, note that the last sum is telescopic.

Robert Z
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