All Questions
250
questions
58
votes
7
answers
25k
views
Is it possible to write a sum as an integral to solve it?
I was wondering, for example,
Can:
$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$
Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.
But ...
- 11.2k
42
votes
6
answers
7k
views
If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third ...
- 621
39
votes
12
answers
90k
views
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? [duplicate]
Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{...
- 433
37
votes
3
answers
3k
views
An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$
$n$ is a positive integer, then
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$
please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.
I want to find a ...
- 7,819
29
votes
2
answers
829
views
How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$
How can we prove the following
$$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$
I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \...
user208998
18
votes
2
answers
865
views
Proving $(λ^d + (1-λ^d)e^{(d-1)s})^{\frac{1}{1-d}}\leq\sum\limits_{n=0}^\infty\frac1{n!}λ^{\frac{(d^n-1)d}{d-1}+n}s^ne^{-λs}$
Question
Let $\lambda \in (0,1), s \in (0,\infty), d \in \{2,3,\dots\}$ and show that in this case the following inequality holds:
$$(\lambda^d + (1-\lambda^d) e^{(d-1)s})^{\frac{1}{1-d}} \leq \sum_{...
- 1,135
17
votes
2
answers
579
views
Is there a way to evaluate analytically the following infinite double sum?
Consider the following double sum
$$
S = \sum_{n=1}^\infty \sum_{m=1}^\infty
\frac{1}{a (2n-1)^2 - b (2m-1)^2} \, ,
$$
where $a$ and $b$ are both positive real numbers given by
\begin{align}
a &= ...
14
votes
2
answers
244
views
Two conjectured identities of sums involving $\sin(x\sin x)/x^2$
While tackling this question, I came up with the conjecture that the following identities hold:
Conjecture. Let $f(x)=\dfrac{\sin(x\sin x)}{x^2}$. Then
$$ \sum_{k=-\infty}^{\infty} f(k\pi+x) = 1 \...
- 170k
13
votes
4
answers
7k
views
Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$
So I have this problem for midterm reviews:
$$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$
I know that you can find the series form of a natural log, as shown here:
$$\ln\left(1-\...
- 427
13
votes
5
answers
1k
views
Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$
I know how to prove that
$$\sum_1^{\infty} \frac{1}{n^2}<2$$ because
$$\sum_1^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}<2$$
But I wanted to prove it using only inequalities. Is there a way to do ...
- 6,756
13
votes
1
answer
692
views
Summation of an Infinite Series: $\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$
I am having trouble proving that
$$\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$$
I know that
$$\frac{2x \ \arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=1}^\infty \frac{(2x)^{2n}}{...
- 11.6k
11
votes
2
answers
689
views
Infinite Series $\sum\limits_{k=1}^{\infty}\frac{k^n}{k!}$
How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$?
for example for $n=6$, we have
$$\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.$$
- 5,626
11
votes
4
answers
2k
views
A power series $\sum_{n = 0}^\infty a_nx^n$ such that $\sum_{n=0}^\infty a_n= +\infty$ but $\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \ne \infty$
Let's consider the power series $\sum_{n = 0}^{\infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $\sum_{n = 0}^{\infty} a_n= +\infty$. Then I would like to find a sequence ...
- 666
11
votes
2
answers
441
views
An Euler type sum: $\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$, where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$
I've been trying to compute the following series for quite a while :
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$$
where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ are the ...
- 1,984
11
votes
3
answers
207
views
Prove that $\lim_{a \to \infty} \sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = 1$.
The above sum (without the $\lim$ notation) is convergent $\forall a \in \Bbb{N}^+$, because:
$$
\sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = \sum_{n=1}^{\infty} \frac{n! n! \dots n!}{n^n n^n \dots n^n}...
- 2,740