# Tagged Questions

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### Convergence of $\sum(n^3\sin^2n)^{-1}$

I saw a while ago in a book by Clifford Pickover, that whether $\displaystyle \sum_{n=1}^\infty\frac1{n^3\sin^2 n}$ converges is open. I would think that the question of its convergence is really ...
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### Is $\sum\limits_{n=0}^\infty x^n / \sqrt{n!}$ positive?

Is $$\sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0$$ for all real $x$? (I think it is.) If so, how would one prove this? (To confirm: This is the power series for $e^x$, except with the ...
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### Nonexistence of boundary between convergent and divergent series?

The following is a FAQ that I sometimes get asked, and it occurred to me that I do not have an answer that I am completely satisfied with. In Rudin's Principles of Mathematical Analysis, following ...
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### The unreasonable effectiveness of Pade approximation

I am trying to get an intuitive feel for why the Pade approximation works so well. Given a truncated Taylor/Maclaurin series it "extrapolates" it beyond the radius of convergence. But what I can't ...
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### Probabilities in a riddle involving axiom of choice

The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question): The Riddle: We assume ...
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### Zeroes of the random Fibonacci sequence

Let X_n be the "random Fibonacci sequence," defined as follows: $X_0 = 0, X_1 = 1$; $X_n = \pm X_{n-1} \pm X_{n-2}$, where the signs are chosen by independent 50/50 coinflips. It is known that |X_n|...
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### Evaluating an infinite sum related to $\sinh$

How can we show the following equation $$\sum_{n\text{ odd}}\frac1{n\sinh(n\pi)}=\frac{\mathrm{ln}2}8\;?$$ I found it in a physics book(David J. Griffiths,'Introduction to electrodynamics',in ...
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### Is this a rational function?

Is $$\sum_{n=1}^{\infty} \frac{z^n}{2^n-1} \in \mathbb{C}(z)\ ?$$ In a slightly different vein, given a sequence of real numbers $\{a_n\}_{n=0}^\infty$, what are some necessary and sufficient ...
The sequence $(a_n)_{n \ge 0}$ satisfies, $a_0 = a_1 = 1$ and the recursion relation: $$a_n = \sum\limits_{k=0}^{[n/2]} \frac{a_k}{(n-2k)!}$$ where, $[x]$ is the nearest integer to $x$ not exceeding ...