Linked Questions

216 votes
26 answers
19k views

There have already been some questions about some undocumented functionality in Mathematica. Such as (please add to these lists!) How can one find undocumented options or option values in Mathematica?...
14 votes
7 answers
3k views

I am trying to compute numerically NSum[(-1)^n/n^3, {n, 1, Infinity}]. Of course, using first Sum would work here, but often it'...
user avatar
12 votes
5 answers
599 views

Revisiting the problem Limit of partial sums involving inverse squares I found another difficulty with Sum[] Consider this sum ...
Dr. Wolfgang Hintze's user avatar
8 votes
3 answers
660 views

Mathematica 11 produces ...
user64494's user avatar
11 votes
3 answers
383 views

I want to show the following identity: $$\sum_{n=-\infty}^{+\infty}\frac{1}{x_n^2}=\frac{1}{5},$$ where $x_n$ are the non-zero solutions of $$\tan(x) = x.$$ I know how to prove the correctness of this ...
Laura Olatex's user avatar
16 votes
1 answer
792 views

In older versions of Mathematica, there was a function called SequenceLimit that allowed taking the limit of a numerical sequence. It is useful for speeding up the ...
Chris Chiasson's user avatar
4 votes
3 answers
2k views

I have an integration to do. I want to integrate $$ \int_0^\infty \sin^2(2\pi t)f(t)\mathrm{d}t $$ where $f(t)$ is known only at discrete values given in an array in the form $\{t_i,f_i\}$ with $i=1\...
jason's user avatar
  • 275
2 votes
2 answers
308 views

$$ \sum _{k=1}^{\infty }\frac{4^kH_{2k}^2}{k^2\binom{2k}{k}}=8G^2+\frac{103}{2}\zeta \left(4\right)-22\operatorname{Li}_4\left(\frac{1}{2}\right)+7\ln ^2\left(2\right)\zeta \left(2\right)-\frac{11}{12}...
Ahamad's user avatar
  • 1
3 votes
0 answers
691 views

I am trying to learn to use Mathematica in an efficient way. Thus, I decided to spend some time on functional programming. I would like to implement the so-called epsilon algorithm, that is used to ...
A. Florio's user avatar
0 votes
1 answer
94 views

How can I solve this limit? thanks you!!!
Gabri's user avatar
  • 31