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 |
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.
10
votes
Solving a limit about sum of series
The function $x \mapsto t^{x^2}$ decreases on $\mathbb{R}_+$, so for every $n \in \mathbb{N}$,
$$t^{(n+1)^2} \le \int_n^{n+1}t^{x^2}dx \le t^{n^2}$$
By summation over $n$,
$$\sum_{n=1}^{\infty}t^{n^2} …
2
votes
Does there exist a continuous function $f(x)$ such that $f(0)=0$ and $0<\lim_{n\to\infty}\pr...
Possible way to find such a function $f$
Since
$$\prod_{k=1}^n \frac{ke}{n} = n!\Big(\frac{e}{n}\Big)^n \sim \sqrt{2\pi n} \text{ as } n \to +\infty,$$
it is sufficient to find a continuous function $ …
1
vote
Accepted
Convergence and roots of alternating periodic infinite series
I prove the convergence of the series.
For $n \ge 1$, let
$$S_n = \sum_{k=1}^n k^{-i\beta-\alpha} \text{ and } S'_n = \sum_{k=1}^n (-1)^ {k-1} k^{-i\beta-\alpha}$$
Then
$$S_{2n}-S'_{2n} = 2 \sum_{k=1} …