All Questions
47
questions
2
votes
0
answers
55
views
Closed form for a series involving the $\Gamma$ and $\zeta$ functions
I was just wondering wether one can derive a closed form for $$\sum_{n=1}^{\infty}\frac{1}{\Gamma\left(\frac{1}{n}\right)\zeta\left(1+\frac{1}{n}\right)}$$
Numerical simulation gives $S=1.20154...$
...
- 1,984
0
votes
1
answer
30
views
How much a variable contributed to the result of some cost function
I have this simple cost function: $\sum_{i=1}^n d_i\times h_i \times a_i$
I wanted to analyze, for example, how much the $a$ component/variable contributed to the final cost function. In other ...
3
votes
4
answers
192
views
How to prove that $1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$?
I'm trying to prove that
$$1+\frac11(1+\frac12(1+\frac13(...(1+\frac1{n-1}(1+\frac1n))...)))=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+...+\frac1{n!}$$
Using induction, suppose that
$$1+\frac11(1+\frac12(...
user547800
1
vote
0
answers
86
views
Is the following inequality True or False? [duplicate]
Ι got a feeling that $$\sum_{x=1}^{\infty}\Big\lvert\sum_{k=0}^{\infty} \frac{x^{2k}}{(k+1)!}(-1)^{k} \Big\rvert \geq \sum_{n=1}^{\infty} \frac{1}{n} $$
because $$\sum_{x=1}^{\infty} \Big\lvert x-\...
- 2,782
1
vote
1
answer
35
views
find a where $\sum_{n=n(a)}^\infty (-1)^n \frac{(2+n^a)}{n}$ converges and absolutely converges
Study $a$ so that the series convergences and absolute convergences
I just know that $a$ is positive
I applied the ratio test, I've found that $0<a<1$
I have no idea for absolute convergence.
...
3
votes
2
answers
82
views
How to find the limit:$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$
How to find the limit:$$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$$
I can't think of any way of this problem
Can someone to evaluate this?
...
- 1,431
2
votes
1
answer
232
views
Summation notation two sums?
$$\sum_{n=1}^{2^k - 1}\frac{1}{n} = \sum_{n=1}^{2-1}\frac{1}{n} + \sum_{n=2}^{2^2 - 1}\frac{1}{n} + \cdots + \sum_{n = 2^{k-1}}^{2^k - 1}\frac{1}{n} = \boxed{\sum_{j = 0}^{k-1}\sum_{n = 2^j}^{2^{j+1} -...
- 513
0
votes
1
answer
72
views
A Finite Summation
Is there an easy way to find the following summation (question created by Priyanshu Mishra on Brilliant.org, but now has been deleted):
$$\sum _{n=2}^{25}
\frac{(n+1)!-n!}{n^{18}+n^{17}+n^{15}}$$
...
user485639
4
votes
2
answers
212
views
Asymptotic of a sum: $\sum_{k=n}^{\infty}{f(k)} = \int_{n}^{\infty}{f(t)dt} +\frac{f(n)}{2}+\mathcal{O}(f'(n))$
Someone told me that the following formula holds for $f$ differentiable and decreasing, with $\lim_{x\rightarrow +\infty}{f(x)}=0$.
$$\sum_{k=n}^{\infty}{f(k)} = \int_{n}^{\infty}{f(t)dt} +\frac{f(...
- 301
1
vote
2
answers
487
views
Finding the limit as $k$ tends to infinity of this sum
$\sum_{i=0}^{\lceil zk \rceil}{k\choose i} p^i(1-p)^{k-i}$
$z, p \in [0,1]$
I am looking to find the limit as $k$ tends to infinity but don't know how I would do this
- 125
3
votes
1
answer
612
views
Changing the order of summation and using a change of variables
I'm looking at the below, and I don't understand how to get the second equality using the change of variables $m=k-l$. I've tried changing the order of summation to get $$\sum_{k=2^j+1}^{2^{j+1}}\sum_{...
- 5,611
3
votes
2
answers
2k
views
Partial sums of $nx^n$
WolframAlpha claims:
$$\sum_{n=0}^m n x^n = \frac{(m x - m - 1) x^{m + 1} + x}{(1 - x)^2} \tag{1}$$
I know that one can differentiate the geometric series to compute $(1)$ when it is a series, i.e. $m=...
- 11.8k
4
votes
2
answers
460
views
What's a closed form for $\sum_{k=0}^n\frac{1}{k+1}\sum_{r=0\\r~is~odd}^k(-1)^r{k\choose r}r^n$
I want to use a closed form of
$$\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is odd}}^k(-1)^r{k\choose r}r^n$$
and
$$\sum_{k=0}^n\frac{1}{k+1}\sum_{\ \ \ r=0\\r\text{ is even}}^k(-1)^r{k\choose ...
- 30.1k
7
votes
2
answers
317
views
Stuck with this limit of a sum: $\lim _{n \to \infty} \left(\frac{a^{n}-b^{n}}{a^{n}+b^{n}}\right)$.
Here's the limit: $$\lim _{n \to \infty} \left(\frac{a^{n}-b^{n}}{a^{n}+b^{n}}\right)$$
The conditions are $b>0$ and $a>0$.
I tried this with the case that $a>b$:
$$\lim _{n \to \infty} \...
- 361
0
votes
2
answers
107
views
Calculate (i.e. express without using infinite sum): $\frac{1}{1!\cdot1}+\frac{2}{2!\cdot2}+\frac{4}{4!\cdot4}+\frac{8}{6!\cdot8}+..$
Calculate (i.e. express without using infinite sum):
$$\frac{1}{1!\cdot1}+\frac{2}{2!\cdot2}+\frac{4}{4!\cdot4}+\frac{8}{6!\cdot8}+..$$
In sum it would be:
$$\sum_{n=0}^{\infty}\frac{2^{n}}{(...
- 1,007