Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with sequences-and-series
Search options not deleted
user 4213
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.
15
votes
Accepted
fibonacci series mod a number
This is really just an expansion of Gerhard's comment. One has the matrix formula
$$\begin{pmatrix}
1&1\\\
1&0
\end{pmatrix}^n=
\begin{pmatrix}
F_{n+1}&F_n\\\
F_n&F_{n-1}
\end{pmatrix}
$$
so the probl …
- 20.8k
8
votes
Accepted
mertens-function in the light of divergent summation - what summation method were best adapted
Well,
$$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$
for $s>1$, so setting $s=0$ should give
$$\sum_{n=1}^\infty\mu(n)=\frac1{\zeta(0)}=-2$$
as $\zeta(0)=-1/2$. :-)
I should add that this i …
- 20.8k