All Questions
41
questions
1
vote
1
answer
60
views
Why does $f^{(k)}(0)$ exists in Rudin's PMA Corollary 8.1?
is plugging $0$ in (6) result to $0^0$?
here is conditions of $8.1$
- 1,961
4
votes
1
answer
89
views
A conjecture involving series with zeta function
Recently, I tried to evaluate a limit proposed by MSE user Black Emperor. In the process of evaluating the limit, I have obtained the following equality.
$$
\lim_{N\rightarrow \infty} \sum_{n=0}^{N-2}{...
- 33.2k
0
votes
0
answers
60
views
Rewriting a sum with a floor function as upper limit
I am having some trouble in rewriting a sum whose upper limit is given in terms of a floor function $\lfloor \cdot \rfloor$. The task is to prove that both sides of the following expression coincide:
$...
- 50.3k
0
votes
0
answers
105
views
Manipulation with the following infinite sum
Calculating some observable, I obtained the following-like converges sum
$$
S = \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\sum_{p=0}^{\min(n,k)} \frac{x^n}{n!} \frac{y^k}{k!} F(p),
$$
where $F$ - some ...
- 121
0
votes
0
answers
65
views
How to find the sum of a power series without knowing the actual power series
How do I find the sum of this series?
$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n-1}} $$
The approach I wanted to use is to find a power series that can become this number series for a certain ...
- 151k
0
votes
1
answer
71
views
Evaluate $\sum_{i=2}^{\infty}{\frac{n}{n^2-1}}x^n$
Evaluate $$\sum_{i=2}^{\infty}{\frac{n}{n^2-1}}x^n$$ using the fact that
$${\frac{n}{n^2-1}} = {\frac{1}{2(n-1)}} + {\frac{1}{2(n+1)}}$$
So far I have proven that the Radius of Convergence is 1 and ...
- 330
1
vote
3
answers
193
views
Computing:$\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}$
I'm learning the subject Power Series and I can't figure out how to find the sum of the series $$\sum_{n=0}^\infty\frac{3^n}{n!(n+3)}$$.
I know that the power series $\sum_{n=0}^\infty\frac{3^n}{n!(n+...
- 3,338
0
votes
2
answers
52
views
What is the Radius of Convergene of $f(z) = \sum_{n=0}^\infty \frac1{4^n}z^{2n+1}$
Here's what I have:
$f(z) = z + \frac14z^3+\frac1{4^2}z^5+...$
So, my coefficients are either $0$ or $\frac1{4^n}$ with $\frac1{4^n}$ being the supremum.
So, $\limsup \limits_{n \to \infty} |c_n|^{\...
- 40.8k
10
votes
2
answers
2k
views
Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$ [duplicate]
Find the sum:
$$\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}{...
- 4,538
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 ...
- 7,693
3
votes
1
answer
2k
views
Find a sum of a convergent series
Let $x_n$ be a sequence that is given by the following recursive formula:
$x_{n+1} = x_n^2 - x_n +1$, where $x_1=a \gt 1$.
Find: $$\sum_{n=1}^{\infty} \frac{1}{x_n}$$
Not sure really how to ...
CommunityBot
- 1
1
vote
1
answer
49
views
Show that $F(x)=\frac{x}{(1-x)^2}-\frac{2x}{(2-x)^2}=\sum_{n=0}^{\infty}n(1-2^{-n})x^n$
I've been working on a recent exercise question where I was asked to show that:
$$F(x)=\frac{x}{(1-x)^2}-\frac{2x}{(2-x)^2}=\sum_{n=0}^{\infty}n(1-2^{-n})x^n$$
Now I cansee that the infinite sum is ...
- 431k
1
vote
4
answers
111
views
Evaluating the sum $\sum\limits_{n=1}^{\infty}\frac{x^{2n}}{(2n+1)!}$
I'm having difficulties with the sum above. My first attempt was to rewrite it like
$$
\sum\limits_{n=1}^{\infty}\frac{x^{2n}}{(2n+1)!}=\frac{1}{x}\sum\limits_{n=1}^{\infty}\frac{x^{2n+1}}{(2n+1)!}
$$
...
- 8,723
0
votes
1
answer
46
views
Sum of a finite series almost like gp
Let $a>1$ and consider the following finite series:
$$
1+\frac{2}{a}+\frac{3}{a^2}+\cdots+\frac{n}{a^{n-1}},
$$
where $n\geq 1$ is a fixed quantity.
Then is the above series uniformly bounded by a ...
- 47.4k
0
votes
2
answers
254
views
Infinite power series sum $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$
Using theorems about differentiation or integration of power series calculate infinite sum of
$$
\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}
$$
The answer should equal to $\frac{\pi}{2\sqrt3}$.
I ...
- 431k