All Questions
53
questions
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
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
6
votes
5
answers
512
views
Infinite Series $1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$
Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$:
$$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$
...
- 61
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$$
But I wanted to prove it using only inequalities. Is there a way to do ...
- 6,756
9
votes
4
answers
456
views
Limit of the series $\sum_{k=1}^\infty \frac{n}{n^2+k^2}.$
I am trying to evaluate $$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$ Now I am aware that clearly $$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{...
- 1,018
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
10
votes
1
answer
382
views
On convergence of series of the generalized mean $\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s.$
Assume that $a_n>0$ such that $\sum_{n=1}^{\infty}a_n $ converges.
Question: For what values of $s\in \Bbb R$ does the following series :
$$ I_s= \sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^...
- 24.2k
8
votes
1
answer
253
views
Proving that $\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s$ converges when $\sum_{n=1}^{\infty}a_n $ converges
Assume that $a_n\ge0$ such that $\sum_{n=1}^{\infty}a_n $ converges, then
show that for every $s>1$ the following series converges too:
$$\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\...
- 24.2k
7
votes
2
answers
349
views
Prove that $\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$
Prove that $$\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$
I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$
And maybe taking it's ...
- 2,646
4
votes
2
answers
396
views
Double sum identity
Consider the following double sum identities
$$\sum_{n=0}^\infty\sum_{m=0}^n a(m,n-m) = \sum_{p=0}^\infty\sum_{q=0}^\infty a(p,q) = \sum_{r=0}^\infty\sum_{s=0}^{\lfloor{r/2}\rfloor} a(s,r-2s)$$
The ...
- 26.9k
3
votes
1
answer
325
views
$\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}$
Hi I am trying to calculate the sum given by
$$
\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}=\ = \sqrt{\frac{\pi}{\alpha}} e^{\beta^2/(4\alpha)} \vartheta_3\big(-\frac{\pi\beta}{2\alpha},e^{-\pi^2/(...
- 9,966
58
votes
7
answers
25k
views
Is it possible to write a sum as an integral to solve it?
I was wondering, for example,
Can:
$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$
Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals.
But ...
- 11.2k
29
votes
2
answers
829
views
How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$
How can we prove the following
$$\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
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