list of mathematical series

{{See also|Series (mathematics)#Examples of numerical series|Summation}}

This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums.

Here, $0^0$ is taken to have the value $1$
$\{x\}$ denotes the fractional part of $x$
$B_n(x)$ is a Bernoulli polynomial.
$B_n$ is a Bernoulli number, and here, $B_1=-\frac{1}{2}.$
$E_n$ is an Euler number.
$\zeta(s)$ is the Riemann zeta function.
$\Gamma(z)$ is the gamma function.
$\psi_n(z)$ is a polygamma function.
$\operatorname{Li}_s(z)$ is a polylogarithm.
$n \choose k$ is binomial coefficient
$\exp(x)$ denotes exponential of $x$

Sums of powers

See Faulhaber's formula.

$\sum_{k=0}^m k^{n-1}=\frac{B_n(m+1)-B_n}{n}$

The first few values are:

$\sum_{k=1}^m k=\frac{m(m+1)}{2}$
$\sum_{k=1}^m k^2=\frac{m(m+1)(2m+1)}{6}=\frac{m^3}{3}+\frac{m^2}{2}+\frac{m}{6}$
$\sum_{k=1}^m k^3$

=\left[\frac{m(m+1)}{2}\right]^2=\frac{m^4}{4}+\frac{m^3}{2}+\frac{m^2}{4}

See zeta constants.

$\zeta(2n)=\sum^{\infty}_{k=1} \frac{1}{k^{2n}}=(-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!}$

The first few values are:

$\zeta(2)=\sum^{\infty}_{k=1} \frac{1}{k^2}=\frac{\pi^2}{6}$ (the Basel problem)
$\zeta(4)=\sum^{\infty}_{k=1} \frac{1}{k^4}=\frac{\pi^4}{90}$
$\zeta(6)=\sum^{\infty}_{k=1} \frac{1}{k^6}=\frac{\pi^6}{945}$

Power series

=Low-order polylogarithms=

Finite sums:

$\sum_{k=m}^{n} z^k = \frac{z^{m}-z^{n+1}}{1-z}$ , (geometric series)
$\sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z}$
$\sum_{k=1}^{n} z^k = \frac{1-z^{n+1}}{1-z}-1 = \frac{z-z^{n+1}}{1-z}$
$\sum_{k=1}^n k z^k = z\frac{1-(n+1)z^n+nz^{n+1}}{(1-z)^2}$
$\sum_{k=1}^n k^2 z^k = z\frac{1+z-(n+1)^2z^n+(2n^2+2n-1)z^{n+1}-n^2z^{n+2}}{(1-z)^3}$
$\sum_{k=0}^n k^m z^k = \left(z \frac{d}{dz}\right)^m \frac{1-z^{n+1}}{1-z}$

Infinite sums, valid for $|z|<1$ (see polylogarithm):

$\operatorname{Li}_n(z)=\sum_{k=1}^{\infty} \frac{z^k}{k^n}$

The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form:

$\frac{\mathrm{d}}{\mathrm{d}z}\operatorname{Li}_n(z)=\frac{\operatorname{Li}_{n-1}(z)}{z}$
$\operatorname{Li}_{1}(z)=\sum_{k=1}^\infty \frac{z^k}{k}=-\ln(1-z)$
$\operatorname{Li}_{0}(z)=\sum_{k=1}^\infty z^k=\frac{z}{1-z}$
$\operatorname{Li}_{-1}(z)=\sum_{k=1}^\infty k z^k=\frac{z}{(1-z)^2}$
$\operatorname{Li}_{-2}(z)=\sum_{k=1}^\infty k^2 z^k=\frac{z(1+z)}{(1-z)^3}$
$\operatorname{Li}_{-3}(z)=\sum_{k=1}^\infty k^3 z^k =\frac{z(1+4z+z^2)}{(1-z)^4}$
$\operatorname{Li}_{-4}(z)=\sum_{k=1}^\infty k^4 z^k =\frac{z(1+z)(1+10z+z^2)}{(1-z)^5}$

=Exponential function=

$\sum_{k=0}^\infty \frac{z^k}{k!} = e^z$
$\sum_{k=0}^\infty k\frac{z^k}{k!} = z e^z$ (cf. mean of Poisson distribution)
$\sum_{k=0}^\infty k^2 \frac{z^k}{k!} = (z + z^2) e^z$ (cf. second moment of Poisson distribution)
$\sum_{k=0}^\infty k^3 \frac{z^k}{k!} = (z + 3z^2 + z^3) e^z$
$\sum_{k=0}^\infty k^4 \frac{z^k}{k!} = (z + 7z^2 + 6z^3 + z^4) e^z$
$\sum_{k=0}^\infty k^n \frac{z^k}{k!} = z \frac{d}{dz} \sum_{k=0}^\infty k^{n-1} \frac{z^k}{k!}\,\! = e^z T_{n}(z)$

where $T_{n}(z)$ is the Touchard polynomials.

=Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship=

$\sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{(2k+1)!}=\sin z$
$\sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)!}=\sinh z$
$\sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(2k)!}=\cos z$
$\sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}=\cosh z$
$\sum_{k=1}^\infty \frac{(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tan z, |z|<\frac{\pi}{2}$
$\sum_{k=1}^\infty \frac{(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tanh z, |z|<\frac{\pi}{2}$
$\sum_{k=0}^\infty \frac{(-1)^k2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\cot z, |z|<\pi$
$\sum_{k=0}^\infty \frac{2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\coth z, |z|<\pi$
$\sum_{k=0}^\infty \frac{(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\csc z, |z|<\pi$
$\sum_{k=0}^\infty \frac{-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\operatorname{csch} z, |z|<\pi$
$\sum_{k=0}^\infty \frac{(-1)^kE_{2k}z^{2k}}{(2k)!}=\operatorname{sech} z, |z|<\frac{\pi}{2}$
$\sum_{k=0}^\infty \frac{E_{2k}z^{2k}}{(2k)!}=\sec z, |z| < \frac{\pi}{2}$
$\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{(2k)!}=\operatorname{ver}z$ (versine)
$\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{2(2k)!}=\operatorname{hav}z$ {{cite web | first=Eric W. | last=Weisstein | author-link=Eric W. Weisstein | title=Haversine | work=MathWorld | publisher=Wolfram Research, Inc. | url=http://mathworld.wolfram.com/Haversine.html | access-date=2015-11-06 |url-status=live |archive-url=https://web.archive.org/web/20050310194740/http://mathworld.wolfram.com/Haversine.html |archive-date=2005-03-10}} (haversine)
$\sum_{k=0}^\infty \frac{(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\arcsin z, |z|\le1$
$\sum_{k=0}^\infty \frac{(-1)^k(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\operatorname{arcsinh} {z}, |z| \le 1$
$\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{2k+1}=\arctan z, |z|<1$
$\sum_{k=0}^\infty \frac{z^{2k+1}}{2k+1}=\operatorname{arctanh} z, |z|<1$
$\ln2+\sum_{k=1}^\infty \frac{(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^2}=\ln\left(1+\sqrt{1+z^2}\right), |z| \le 1$
$\sum_{k=2}^\infty \left( k \cdot \operatorname{arctanh}\left(\frac{1}{k}\right) - 1 \right) = \frac{3-\ln(4 \pi)}{2}$

=Modified-factorial denominators=

$\sum^{\infty}_{k=0} \frac{(4k)!}{2^{4k} \sqrt{2} (2k)! (2k+1)!} z^k = \sqrt{\frac{1-\sqrt{1-z}}{z}}, |z|<1$ {{cite book|url=http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf|title=generatingfunctionology|first=Herbert R.|last=Wilf|date=1994|publisher=Academic Press, Inc}}
$\sum^{\infty}_{k=0} \frac{2^{2k} (k!)^2}{(k+1) (2k+1)!} z^{2k+2} = \left(\arcsin{z}\right)^2, |z|\le1$
$\sum^{\infty}_{n=0} \frac{\prod_{k=0}^{n-1}(4k^2+\alpha^2)}{(2n)!} z^{2n} + \sum^{\infty}_{n=0} \frac{\alpha \prod_{k=0}^{n-1}[(2k+1)^2+\alpha^2]}{(2n+1)!} z^{2n+1} = e^{\alpha \arcsin{z}}, |z|\le1$

= Binomial coefficients =

$(1+z)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} z^k, |z|<1$ (see {{slink|Binomial theorem|Newton's generalized binomial theorem}})
{{cite web|url=http://www.tug.org/texshowcase/cheat.pdf|title=Theoretical computer science cheat sheet}} $\sum_{k=0}^\infty {{\alpha+k-1} \choose k} z^k = \frac{1}{(1-z)^\alpha}, |z|<1$
$\sum_{k=0}^\infty \frac{1}{k+1}{2k \choose k} z^k = \frac{1-\sqrt{1-4z}}{2z}, |z|\leq\frac{1}{4}$ , generating function of the Catalan numbers
$\sum_{k=0}^\infty {2k \choose k} z^k = \frac{1}{\sqrt{1-4z}}, |z|<\frac{1}{4}$ , generating function of the Central binomial coefficients
$\sum_{k=0}^\infty {2k + \alpha \choose k} z^k = \frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^\alpha, |z|<\frac{1}{4}$

=Harmonic numbers=

(See harmonic numbers, themselves defined $H_n = \sum_{j=1}^{n} \frac{1}{j}$ , and $H(x)$ generalized to the real numbers)

$\sum_{k=1}^\infty H_k z^k = \frac{-\ln(1-z)}{1-z}, |z|<1$
$\sum_{k=1}^\infty \frac{H_k}{k+1} z^{k+1} = \frac{1}{2}\left[\ln(1-z)\right]^2, \qquad |z|<1$
$\sum_{k=1}^\infty \frac{(-1)^{k-1} H_{2k}}{2k+1} z^{2k+1} = \frac{1}{2} \arctan{z} \log{(1+z^2)}, \qquad |z|<1$
$\sum_{n=0}^\infty \sum_{k=0}^{2n} \frac{(-1)^k}{2k+1} \frac{z^{4n+2}}{4n+2} = \frac{1}{4} \arctan{z} \log{\frac{1+z}{1-z}},\qquad |z|<1$
$\sum_{n=0}^\infty \frac{x^2}{n^2(n+x)} = x\frac{\pi^2}{6} - H(x)$

Binomial coefficients

$\sum_{k=0}^n {n \choose k} = 2^n$
$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$
$\sum_{k=0}^n (-1)^k {n \choose k} = 0, \text{ where }n\geq 1$
$\sum_{k=0}^n {k \choose m} = { n+1 \choose m+1 }$
$\sum_{k=0}^n {m+k-1 \choose k} = { n+m \choose n }$ (see Multiset)
$\sum_{k=0}^n {\alpha \choose k}{\beta \choose n-k} = {\alpha+\beta \choose n}, \text{where} \ \alpha + \beta \geq n$ (see Vandermonde identity)
$\sum_{A \ \in \ \mathcal{P}(E)} 1 = 2^n \text{, where }E\text{ is a finite set, and card(}E\text{) = n}$
$\sum_{\begin{cases} (A,\ B) \ \in \ (\mathcal{P}(E))^2 \\ A \ \subset\ B \end{cases}} 1 = 3^n\text{, where }E\text{ is a finite set, and card(}E\text{) = n}$
$\sum_{A \ \in \ \mathcal{P}(E)} card(A) = n2^{n-1} \text{, where }E\text{ is a finite set, and card(}E\text{) = n}$

Trigonometric functions

Sums of sines and cosines arise in Fourier series.

$\sum_{k=1}^\infty \frac{\cos(k\theta)}{k}=-\frac{1}{2}\ln(2-2\cos\theta)=-\ln \left(2\sin\frac{\theta}{2} \right), 0<\theta<2\pi$
$\sum_{k=1}^\infty \frac{\sin(k\theta)}{k}=\frac{\pi-\theta}{2}, 0<\theta<2\pi$
$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\cos(k\theta)=\frac{1}{2}\ln(2+2\cos\theta)=\ln \left(2\cos\frac{\theta}{2}\right), 0\leq\theta<\pi$
$\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\sin(k\theta)=\frac{\theta}{2}, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sum_{k=1}^\infty \frac{\cos(2k\theta)}{2k}=-\frac{1}{2}\ln(2\sin\theta), 0<\theta<\pi$
$\sum_{k=1}^\infty \frac{\sin(2k\theta)}{2k}=\frac{\pi-2\theta}{4}, 0<\theta<\pi$
$\sum_{k=0}^\infty \frac{\cos[(2k+1)\theta]}{2k+1}=\frac{1}{2}\ln \left(\cot\frac{\theta}{2}\right), 0<\theta<\pi$
$\sum_{k=0}^\infty \frac{\sin[(2k+1)\theta]}{2k+1}=\frac{\pi}{4}, 0<\theta<\pi$ ,

Calculate the Fourier expansion of the function $f(x)=\frac\pi4$ on the interval $0 :$

$\frac\pi4=\sum_{n=0}^\infty c_n\sin[nx]+d_n\cos[nx]$

$\Rightarrow \begin{cases}c_n=\begin{cases}\frac1n\quad (n \text{ odd})\\
0\quad (n \text{ even})\end{cases}\\ d_n=0\quad (\forall n)\end{cases}$

$\sum_{k=1}^\infty \frac{\sin(2 \pi k x)}{k}= \pi \left(\dfrac{1}{2} - \{x\}\right), \ x \in \mathbb{R}$
$\sum\limits_{k=1}^{\infty} \frac{\sin \left(2\pi kx \right)}{k^{2n-1}} = (-1)^{n}\frac{(2\pi)^{2n-1}}{2(2n-1)!} B_{2n-1}(\{x\}), \ x \in \mathbb{R}, \ n \in \mathbb{N}$
$\sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}} = (-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!} B_{2n}(\{x\}), \ x \in \mathbb{R}, \ n \in \mathbb{N}$
$B_n(x)=-\frac{n!}{2^{n-1}\pi^n}\sum_{k=1}^\infty \frac{1}{k^n}\cos\left(2\pi kx-\frac{\pi n}{2}\right), 0 {{cite web|url=http://functions.wolfram.com/Polynomials/BernoulliB2/06/02/|title=Bernoulli polynomials: Series representations (subsection 06/02)|accessdate=2 June 2011|website=Wolfram Research}}$
$\sum_{k=0}^n \sin(\theta+k\alpha)=\frac{\sin\frac{(n+1)\alpha}{2}\sin(\theta+\frac{n\alpha}{2})}{\sin\frac{\alpha}{2}}$
$\sum_{k=0}^n \cos(\theta+k\alpha)=\frac{\sin\frac{(n+1)\alpha}{2}\cos(\theta+\frac{n\alpha}{2})}{\sin\frac{\alpha}{2}}$
$\sum_{k=1}^{n-1} \sin\frac{\pi k}{n}=\cot\frac{\pi}{2n}$
$\sum_{k=1}^{n-1} \sin\frac{2\pi k}{n}=0$
$\sum_{k=0}^{n-1} \csc^2\left(\theta+\frac{\pi k}{n}\right)=n^2\csc^2(n\theta)$ {{cite web|last=Hofbauer|first=Josef|title=A simple proof of 1 + 1/2² + 1/3² + ··· = {{pi}}²/6 and related identities|url=http://homepage.univie.ac.at/josef.hofbauer/02amm.pdf|accessdate=2 June 2011}}
$\sum_{k=1}^{n-1} \csc^2\frac{\pi k}{n}=\frac{n^2-1}{3}$
$\sum_{k=1}^{n-1} \csc^4\frac{\pi k}{n}=\frac{n^4+10n^2-11}{45}$

Rational functions

$\sum_{n=a+1}^{\infty} \frac{a}{n^2 - a^2} = \frac{1}{2} H_{2a}$

{{cite web|last1=Sondow|first1=Jonathan|last2=Weisstein|first2=Eric W.|title=Riemann Zeta Function (eq. 52)|website=MathWorld—A Wolfram Web Resource|url=http://mathworld.wolfram.com/RiemannZetaFunction.html}}

$\sum_{n=0}^\infty\frac{1}{n^2+a^2}=\frac{1+a\pi\coth (a\pi)}{2a^2}$
$\sum_{n=0}^\infty\frac{(-1)^n}{n^2+a^2} = \frac{1 + a\pi \; \text{csch}(a\pi)}{2a^2}$
$\sum_{n=0}^\infty\frac{(2n+1)(-1)^n}{(2n+1)^2+a^2}= \frac{\pi}{4} \text{sech} \left( \frac{a \pi}{2} \right)$
$\displaystyle \sum_{n=0}^\infty \frac {1}{n^4+4a^4} = \dfrac{1}{8a^4}+\dfrac{\pi(\sinh(2\pi a)+\sin(2\pi a))}{8a^3(\cosh(2\pi a)-\cos(2\pi a))}$
An infinite series of any rational function of $n$ can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition,{{cite book|chapter-url=http://people.math.sfu.ca/~cbm/aands/page_260.htm|title=Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables|first1=Milton|last1=Abramowitz|author-link1=Milton Abramowitz|first2=Irene|last2=Stegun|author-link2=Irene Stegun|date=1964|isbn=0-486-61272-4|page=[https://archive.org/details/handbookofmathe000abra/page/260 260]|chapter=6.4 Polygamma functions|publisher=Courier Corporation }} as explained here. This fact can also be applied to finite series of rational functions, allowing the result to be computed in constant time even when the series contains a large number of terms.

Exponential function

$\displaystyle \dfrac{1}{\sqrt{p}}\sum_{n=0}^{p-1}\exp \left(\frac{2\pi i n^2 q}{p} \right)=\dfrac{e^{\pi i/4}}{\sqrt{2q}}\sum_{n=0}^{2q-1}\exp \left(-\frac{\pi i n^2 p}{2q} \right)$ (see the Landsberg–Schaar relation)
$\displaystyle \sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$

== Numeric series ==

These numeric series can be found by plugging in numbers from the series listed above.

=Alternating harmonic series=

$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\ln 2$
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{2k-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots=\frac{\pi}{4}$

=Sum of reciprocal of factorials=

$\sum^{\infty}_{k=0} \frac{1}{k!}=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots=e$
$\sum^{\infty}_{k=0} \frac{1}{(2k)!}=\frac{1}{0!}+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\frac{1}{8!}+\cdots=\frac{1}{2}\left(e+\frac{1}{e}\right)=\cosh 1$
$\sum^{\infty}_{k=0} \frac{1}{(3k)!}=\frac{1}{0!}+\frac{1}{3!}+\frac{1}{6!}+\frac{1}{9!}+\frac{1}{12!}+\cdots=\frac{1}{3}\left(e+\frac{2}{\sqrt{e}}\cos \frac{\sqrt{3}}{2}\right)$
$\sum^{\infty}_{k=0} \frac{1}{(4k)!}=\frac{1}{0!}+\frac{1}{4!}+\frac{1}{8!}+\frac{1}{12!}+\frac{1}{16!}+\cdots=\frac{1}{2}\left(\cos 1+\cosh 1\right)$

=Trigonometry and π=

$\sum^{\infty}_{k=0} \frac{(-1)^k}{(2k+1)!}=\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}+\cdots=\sin 1$
$\sum^{\infty}_{k=0} \frac{(-1)^k}{(2k)!}=\frac{1}{0!}-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}+\cdots=\cos 1$
$\sum^{\infty}_{k=1} \frac{1}{k^2+1}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\cdots=\frac{1}{2}(\pi \coth \pi - 1)$
$\sum^{\infty}_{k=1} \frac{(-1)^k}{k^2+1}=-\frac{1}{2}+\frac{1}{5}-\frac{1}{10}+\frac{1}{17}+\cdots=\frac{1}{2}(\pi \operatorname{csch} \pi - 1)$
$3 + \frac{4}{2\times3\times4} - \frac{4}{4\times5\times6} + \frac{4}{6\times7\times8} - \frac{4}{8\times9\times10} + \cdots = \pi$

===Reciprocal of tetrahedral numbers===

$\sum^{\infty}_{k=1} \frac{1}{Te_k}=\frac{1}{1}+\frac{1}{4}+\frac{1}{10}+\frac{1}{20}+\frac{1}{35}+\cdots=\frac{3}{2}$

Where $Te_n=\sum^{n}_{k=1} T_k$

=Exponential and logarithms=

$\sum^{\infty}_{k=0} \frac{1}{(2k+1)(2k+2)}=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+\frac{1}{7\times 8}+\frac{1}{9\times 10}+\cdots=\ln 2$
$\sum^{\infty}_{k=1} \frac{1}{2^kk}=\frac{1}{2}+\frac{1}{8}+\frac{1}{24}+\frac{1}{64}+\frac{1}{160}+\cdots=\ln 2$
$\sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{2^kk}+\sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{3^kk}=\Bigg(\frac{1}{2}+\frac{1}{3}\Bigg)-\Bigg(\frac{1}{8}+\frac{1}{18}\Bigg)+\Bigg(\frac{1}{24}+\frac{1}{81}\Bigg)-\Bigg(\frac{1}{64}+\frac{1}{324}\Bigg)+\cdots=\ln 2$
$\sum^{\infty}_{k=1} \frac{1}{3^kk}+\sum^{\infty}_{k=1} \frac{1}{4^kk}=\Bigg(\frac{1}{3}+\frac{1}{4}\Bigg)+\Bigg(\frac{1}{18}+\frac{1}{32}\Bigg)+\Bigg(\frac{1}{81}+\frac{1}{192}\Bigg)+\Bigg(\frac{1}{324}+\frac{1}{1024}\Bigg)+\cdots=\ln 2$
$\sum^{\infty}_{k=1} \frac{1}{n^kk}=\ln\left(\frac{n}{n-1}\right)$ , that is $\forall n>1$

Notes

References

Many books with a list of integrals also have a list of series.

Series