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for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.

8 votes

closed form for an alternating cosecant sum

The sum can be rewritten as an integral, by means of $$\frac{1}{\sin (\pi j/n)}=\frac{n}{\pi}\int_0^\infty\frac{s^{j-1}}{ s^n+1}\,ds,\;\;0<j<n,$$ $$\Rightarrow \sum_{j=1}^{n-1}\frac{(-1)^j}{\sin (\pi …
Carlo Beenakker's user avatar
12 votes
Accepted

Erroneous Wolfram result for $\sum_{k=1}^\infty (k^3 + a^3)^{-1}$, looking for correct formula

I think the statement in the OP that $W_2(a)$ and $W_3(a)$ remain bounded when $a\rightarrow 0$ is mistaken, so that there is no inconsistency with the Mathematica result. The three roots of $(w+1)^3+ …
Carlo Beenakker's user avatar
1 vote

Sum of an infinite series

A closed-form answer is not forthcoming, but for $c\lesssim 1$ the small-$c$ approximation is quite accurate: $$S_0=\sum_{n=1}^\infty \frac{x^n}{n!}(1+c/n)=e^x-1+c \bigl[\text{Chi}(x)+\text{Shi}(x)-\l …
Carlo Beenakker's user avatar
3 votes
Accepted

how to prove identity for nth derivative of $(\text{arctanh}(x))^j$?

The derivative of the arctanh can be evaluated in terms of the Stirling numbers by following the suggestion of Tyma Gaidash. Start from the generating function of the Stirling numbers of the first kin …
Carlo Beenakker's user avatar
15 votes
Accepted

Closed form for $\sum_{n=0}^\infty \frac1{2^{2^n}}$?

If you allow for a named number to be a closed form representation, the answer is "yes". $\sum_{n=0}^\infty (1/2)^{2^n}$ is known as the Kempner number [1], a transcendental number [2]. More generally …
Carlo Beenakker's user avatar
3 votes
Accepted

Does any such family of functions exist?

Here is a real solution for $k=2$: Take $f_1(x)=x$ and $$f_2(x)=\arccos\left( \sqrt{\tfrac{3}{4}- \cos ^4(x/2)} \operatorname{cotan} (x/2)-\tfrac{1}{2}\cos x\right).$$ Then $$\cos f_1(x)+\cos f_2(x)=\ …
Carlo Beenakker's user avatar
4 votes

Does anyone remember what happened to the experimental search for polynomial identities for ...

To follow up on my comment, the specific experimental search for polynomial identities that relate powers of $\pi$ and values of the Riemann zeta function, is in Experimental Evaluation of Euler Sums …
Carlo Beenakker's user avatar
2 votes

How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n...

The formula in the OP is a hypergeometric function identity, $$\frac{1}{41}\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=$$ $$\qquad= \,_4F_3\left(\frac{1}{2},\frac …
Carlo Beenakker's user avatar
12 votes
Accepted

Finding $\sum_i x_i$ given $\{\sum_i x_i^{2n}\}_{n\in \mathbb{N}}$

If you plot $\log f_n$ versus $n$, with $f_n=\sum_i x_i^{2n}$, then the asymptotic slope for large $n$ will give you the largest of the $x_i$; subtracting that contribution from $f_n$ and repeating wi …
Carlo Beenakker's user avatar
5 votes

Asymptotic behavior of the "Cauchy square" series

You can reduce the evaluation of $S_n$ to a quadrature by means of the Abel-Plana formula, $$S_n=\sum _{s=1}^{n-1} g(s)=\int_1^{n-1}g(s)\,ds+\tfrac{1}{2}g(1)+\tfrac{1}{2}g(n-1)$$ $$\qquad\qquad -\,2\o …
Carlo Beenakker's user avatar
6 votes

How to evaluate inverse Laplace transform of $e^{- \sqrt{s}} $ using series?

Let me first look at a simpler example, instead of the square root consider the inverse Laplace transform of $e^{-s}$. If you write the series expansion and invert term by term you obtain $$L^{-1}_s\l …
Carlo Beenakker's user avatar
5 votes

Binomial series

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left …
Carlo Beenakker's user avatar
1 vote

If $\inf\{b\in\mathbb{R}\mid\sum_{n=1}^{\infty}e^{-ax_n-by_n}<+\infty\}=1-a$ for all $a\in [...

Q: Any article studying the series $\sum_{n=1}^{+\infty}e^{-ax_n-by_n}$? A: This is the grand canonical partitition sum studied in statistical mechanics, in the form $$Z=\sum_k e^{-\beta E_k+\beta\mu …
Carlo Beenakker's user avatar
5 votes

Four new series for $\pi$ and related identities involving harmonic numbers

a bit long for a comment. The "four new series for $\pi$" are examples of relationships between hypergeometric functions $_pF_{p-1}$ with rational arguments, for example, the first series is equivalen …
Carlo Beenakker's user avatar
4 votes
Accepted

Abel–Plana formula with fractional offset

I eventually did find a published derivation of the fractional-offset (2) of the Abel-Plana formula, in A Generalized Mode Summation Formula of the Zero-Point Energy in a Cavity by Norio Inui (2003): …
Carlo Beenakker's user avatar

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