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Results tagged with sequences-and-series
Search options questions only
not deleted
user 66131
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.
46
votes
5
answers
4k
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Fibonacci series captures Euler $e=2.718\dots$
The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question m …
- 43.2k
23
votes
4
answers
2k
views
Identity for an infinite product
Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes".
QUESTION. Is this true?
$$\prod_{n\geq1}(1+x^{2n-1})^{24} - \prod_{n\geq1}(1-x^{ …
- 43.2k
18
votes
5
answers
3k
views
Bernoulli sum meets golden number
Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the golden ratio.
I encountered the following infinite sum and would like to ask:
Question. Is this true? If so, any pr …
- 43.2k
16
votes
4
answers
2k
views
What can be said about this double sum?
Question. Can this number be expressed in terms of classical values?
$$\sum_{n,m=1}^{\infty}\frac1{(n^2+m^2)^{\frac32}}=1.056348517615643291\dots$$
UPDATE. I'm encouraged by Noam, Kevin and Igo …
- 43.2k
15
votes
4
answers
3k
views
No Tonelli or Fubini
Whenever we can interchange summation (perhaps due to Tonelli-Fubini), good things happen. Otherwise, one has to struggle evaluating double sums in just one way, because the alternative results in a d …
14
votes
4
answers
2k
views
Integrality of a sequence formed by sums
Consider the following sequence defined as a sum
$$a_n=\sum_{k=0}^{n-1}\frac{3^{3n-3k-1}\,(7k+8)\,(3k+1)!}{2^{2n-2k}\,k!\,(2k+3)!}.$$
QUESTION. For $n\geq1$, is the sequence of rational numbers $a_n$ …
- 43.2k
13
votes
3
answers
1k
views
iterated harmonic numbers vs Riemann zeta
Define the $m$-th iterated harmonic sums in the manner: $\bar{H}_0(n):=1$ and for
$m\geq1$ by
$$\bar{H}_m(n):=\sum_{k=1}^n\frac{\bar{H}_{m-1}(k)}k.$$
For example, $\bar{H}_1(n)=\sum_{k=1}^n\frac1k$ …
- 43.2k
12
votes
2
answers
1k
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An interesting identity: in search of a proof -Part I
I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS.
Question. Can you show that
$$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+ …
- 43.2k
12
votes
1
answer
618
views
Determinants: periodic entries $0,1,2,3$
Consider an $n\times n$ matrix $M_n$ where the sequence
$$\{1,2,3,\dots,n^2\} \mod 4=\{1,2,3,0,1,2,3,\dots\}$$ forms a clock-wise spiral, in that given order. For example,
$$M_4=\begin{bmatrix} 1&2&3 …
- 43.2k
12
votes
1
answer
391
views
Looking for a "clever" argument for a $q$-series identity
Consider the below $q$-series identity. One of the things I like about this expansion is how nicely the difference on the left hand side factors to the right hand side of the equation.
$$\prod_{k\geq1 …
- 43.2k
11
votes
3
answers
1k
views
Integrality of a binomial sum
The following sequence appears to be always an integer, experimentally.
QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers?
$$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2 …
- 43.2k
11
votes
2
answers
1k
views
Two divergent series conspiring?
Consider the sequence $a_n=2^{2n}\binom{2n}n^{-1}$. Stirling's approximation shows that $a_n\sim \sqrt{\pi n}$, thus
$$\sum_{n\geq0}\frac{\pi}{2a_n}\qquad \text{and} \qquad
\sum_{n\geq0}\frac{a_n}{2n+ …
- 43.2k
10
votes
2
answers
308
views
Denominators of certain Laurent polynomials
Consider the following somos-like sequence
$$x_n=\frac{x_{n-1}^2+x_{n-2}^2}{x_{n-3}}.$$
It's known that $x_n$ is a Laurent polynomial in $x_0, x_1$ and $x_2$. I got interested in the denominators of t …
- 43.2k
10
votes
4
answers
1k
views
Adventure with infinite series, a curiosity
It is easily verifiable that
$$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$
It is not that difficult to get
$$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$ …
- 43.2k
10
votes
2
answers
886
views
An attempt to generalize the previous inequality
In my previous MO question, the inequality was about a specific series and nicely answered by Cherng-tiao Perng. After testing with a few more numerical infinite sums, I came to realize that perhaps m …
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