All Questions
Tagged with summation power-series
362
questions
60
votes
11
answers
127k
views
The idea behind the sum of powers of 2
I know that the sum of powers of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.
For example, sum of n numbers is $\...
- 721
33
votes
3
answers
4k
views
Sum of Squares of Harmonic Numbers
Let $H_n$ be the $n^{th}$ harmonic number,
$$ H_n = \sum_{i=1}^{n} \frac{1}{i}
$$
Question: Calculate the following
$$\sum_{j=1}^{n} H_j^2.$$
I have attempted a generating function approach but ...
- 331
28
votes
6
answers
4k
views
Sum of the form $r+r^2+r^4+\dots+r^{2^k} = \sum_{i=1}^k r^{2^i}$
I am wondering if there exists any formula for the following power series :
$$S = r + r^2 + r^4 + r^8 + r^{16} + r^{32} + ...... + r^{2^k}$$
Is there any way to calculate the sum of above series (if ...
- 281
19
votes
2
answers
757
views
Closed form of $\sum_{n = 1}^{\infty} \frac{n^{n - k}}{e^{n} \cdot n!}$
When seeing this question I noticed that
$$
\sum_{n = 1}^{\infty} \frac{n^{n - 2}}{e^{n} \cdot n!}
= \frac{1}{2}.
$$
I don't know how to show this, I tried finding a power series that matches that but ...
- 4,858
13
votes
4
answers
727
views
How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$
We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this sum.
$$
\sum_{n=...
user335850
12
votes
5
answers
301
views
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$?
Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase ...
- 13.2k
11
votes
4
answers
2k
views
A power series $\sum_{n = 0}^\infty a_nx^n$ such that $\sum_{n=0}^\infty a_n= +\infty$ but $\lim_{x \to 1} \sum_{n = 0}^\infty a_nx^n \ne \infty$
Let's consider the power series $\sum_{n = 0}^{\infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $\sum_{n = 0}^{\infty} a_n= +\infty$. Then I would like to find a sequence ...
- 666
10
votes
2
answers
295
views
Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$
Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
- 697
10
votes
2
answers
2k
views
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer.
Given
$$(1+x)^n= \binom {n}{0} + \binom{n}{1} x+ \binom{n}{2} x^2+ \cdots + \binom {n}{n} x^n.$$
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive ...
- 8,892
10
votes
4
answers
973
views
Prove $\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$
I found through simulations that
$$\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$$
Is there any proof of this? I've tried to solve it by:
Induction, but it gets too messy.
Binomial ...
- 12.1k
10
votes
2
answers
2k
views
Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$ [duplicate]
Find the sum:
$$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$
My try:
I played a bit with the coefficient to make it look easier/familiar:
First attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{...
- 877
9
votes
3
answers
12k
views
Formula for finite power series
Are there any formula for result of following power series?
$$0\leq q\leq 1$$
$$
\sum_{n=a}^b q^n
$$
- 303
9
votes
1
answer
1k
views
Proving $\pi=(27S-36)/(8\sqrt{3})$, where $S=\sum_{n=0}^\infty\frac{\left(\left\lfloor\frac{n}{2}\right\rfloor!\right)^2}{n!}$ [closed]
I have to prove that:
$$\pi=\frac{27S-36}{8\sqrt{3}}$$
where I know that $$S=\sum_{n=0}^\infty\frac{\left(\left\lfloor\frac{n}{2}\right\rfloor!\right)^2}{n!}$$
Where do I get started?
- 125
8
votes
7
answers
444
views
Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$
This sum $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that $\,\,\displaystyle \sum_{n=1}^\infty ...
- 35.9k
8
votes
1
answer
159
views
Summation calculus: $\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$
How can I solve this?
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}}$$
Actually I tried many direction, but failed.
Please give me some right direction.
$$\sum_{k=1}^n \frac{2^{2^{k-1}}}{1-2^{2^k}} = \...
- 3,433