# Tagged Questions

For questions about recurrence relations, convergence tests, and identifying sequences. For questions on finite sums use the (summation) instead.

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### Why is this series of square root of twos equal $\pi$?

Wikipedia claims this but only cites an offline proof: $$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$ for $n$ square roots and one minus sign. The formula is not the "usual" one, ...
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### Finding the sum- $x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

If $S = x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$ Find S. Note:This is not a GP series.The powers are in GP. My Attempts so far: 1)If $S(x)=x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$ Then $$S(x)-S(x^{2})=x$$ ...
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### Proving the inequality $\frac{\log (1)}{1!}+\frac{\log ^2(2)}{2!}+\frac{\log^3(3)}{3!}+\cdots> \frac{\pi }{4}$

How to prove this inequality? $$\frac{\log (1)}{1!}+\frac{\log ^2(2)}{2!}+\frac{\log^3(3)}{3!}+\cdots> \frac{\pi }{4}$$ The left side looks vaguely like the series for $\exp(x)$: the terms ...
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### An integral identity from Ramanujan's notebooks

Browsing through Ramanujan's notebooks, I found the following identity, without proof of course (Notebook 1, p. 130): In other words (took me a while to realize that the lower integration bound is ...
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### How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$?

I'd like to find out why \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6 \end{align} I tried to rewrite it into a geometric series \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=...
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### Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$

How can we find the exact value of the infinite sum $$\displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}?$$ This problem appears in: T. Andreescu, T. Radulescu & V....
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### This infinitely nested root gives me two answers $\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{…}}}}}}$

I am trying to evaluate $$\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}$$ where those numbers inside roots are $$a_{n+1}=\frac{a_n^2}{2}$$ And I found two ways to solve ...
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### Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$

I discovered the following conjecture by evaluating the integral numerically and then using some inverse symbolic calculation methods to find a possible closed form: \int_0^\infty\frac{\ln x}{\sqrt{...