All Questions
9
questions
0
votes
0
answers
34
views
Variants of geometric sum formula
I know there are closed formulas for sums of the form $\sum_{k=0}^n k^sr^k$
and $\sum_{k=0}^n r^{2n}$ or $\sum_{k=0}^n r^{2n+1}$.
(See https://en.wikipedia.org/wiki/Geometric_series#Sum)
From Sum of ...
3
votes
2
answers
60
views
Closed form expression for sequence of values created by differently signed series
Consider a sequence of terms of powers of $m\in\mathbb{R}$ as
$$
M_n = m^0, m^1, m^2, m^3, \ldots, m^n
$$
and create a set that contains all the values of the various signed combinations of these ...
-1
votes
1
answer
71
views
Closed form for a series involving exponential functions
Is there a closed form for the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\mathrm{e}^{-n \theta}$ ?
The series is convergent for all real values of $\theta$ as $\lim_{n \to \infty} \frac{\mathrm{e}...
0
votes
0
answers
95
views
Closed form for $Q_{m}(x)=\sum _{n=0}^{\infty } \frac{ (-1)^n}{n+m}\frac{x^{n}}{n!}$
I've just stumbled upon this
$$
Q_{m}(x)=\sum _{n=0}^{\infty } \frac{ (-1)^n}{n+m}\frac{x^{n}}{n!}
$$
and i'd like to know if it has a closed form.
Note that $m \geq1 $ is an integer.
Thanks.
1
vote
1
answer
66
views
Values for which this sum can be defined in terms of known constants in a closed form
I'm interested in the sum,
$$\sum_{n=1}^\infty\frac{\zeta(2n)\Gamma(2n)}{\Gamma(2k+2n+2)}x^{2n}$$
Otherwise written as
$$\sum_{n=1}^\infty\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+2k+1)}x^{2n}$$
I am ...
2
votes
2
answers
219
views
Summation of alternating series, Mercator series: $\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}$
I am struggling with solving sum of this alternate series:
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)}\
$$
I know that:
$$
\log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \cdot x^n\
$$
But ...
3
votes
1
answer
224
views
Closed form of the sum of the product of three binomial coefficients
I encountered with this kind of series from the calculation in quantum optics:
$$\sum_{n,m=0}^\infty \sum_{k,l=0}^{\min(n,m)}\binom{n}{k}\binom{m}{l}\binom{n+m-k-l}{m-k}A^{n+m}B^kC^l$$
Provided that ...
2
votes
0
answers
63
views
How to manipulate this summation in the easiest way possible?
$$
D =
\sum_{k=c}^{n}\sum_{j=0}^{k-c}[{k-c \choose j}\ln^{k-c-j}(g(x))[\ln(g) f'(x) f_c^{(j)} X_{n,k(f\rightarrow g)^c} + f_{c}^{(j)} X_{n,k(f \rightarrow g)^{c}}' + \frac{d}{dx}[f_c^{(j)}] X_{n,k(f \...
0
votes
1
answer
171
views
Close form of a power series starting at $n=2$
This is the power series I am looking at $\sum_{n=2}^{\infty}{n(n-1)z^n}$. I want to find the closed form of this power series.
This is my approach, if I divide the power series by $z^2$, then I ...