All Questions
1,809
questions
6
votes
4
answers
241
views
Evaluate $\int_{0}^{1}\{1/x\}^2\,dx$
Evaluate
$$\displaystyle{\int_{0}^{1}\{1/x\}^2\,dx}$$
Where {•} is fractional part
My work
$$\displaystyle{\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \sum\limits_{n = 1}^\infty {\...
user1235604
2
votes
1
answer
78
views
$\frac{(1+x)^n}{(1-x)^3}=a_{0}+a_{1}x+a_{2}x^2+\cdots$ show that ${a_{0}+\cdots+a_{n-1}=\frac{n(n+2)(n+7)2^{n-4}}{3}}$
$$\displaystyle{\frac{(1+x)^n}{(1-x)^3}=a_{0}+a_{1}x+a_{2}x^2+\cdots}$$, show that $$\displaystyle{a_{0}+\cdots+a_{n-1}=\frac{n(n+2)(n+7)2^{n-4}}{3}}$$
When i gave this problem to my friends they said ...
user1235604
2
votes
1
answer
212
views
Calculate the value $\lim_{n\to \infty}\frac{\sum_{j=1}^n \sum_{k=1}^n k^{1/k^j}}{\sqrt[n]{(\sum_{j=1}^n j!)\sum_{j=1}^n j^n}}$
As in title, I want to calculate the following value $$\lim_{n\to \infty}\frac{\sum_{j=1}^n \sum_{k=1}^n k^{1/k^j}}{\sqrt[n]{(\sum_{j=1}^n j!)\sum_{j=1}^n j^n}}.$$
Here is my attempt:
Since $\sum_{j=...
- 312
-1
votes
2
answers
96
views
Evaluate $\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} - \frac{1}{2} \sum_ {m=1}^{\infty} \frac{(-1)^m \mathcal{H}_m}{2m+1}$ [duplicate]
Let's declare $\mathcal{G}$ is constant of Catalanand the $\mathcal{H}_m-st$ mharmonic term. Let it be shown that:
$$\displaystyle{\sum_{m=1}^{\infty} \frac{(-1)^m \mathcal{H}_{2m} }{2m+1} -\frac{1}{2}...
1
vote
2
answers
105
views
Evaluate $\sum_{n=1}^{\infty} (-1)^{n+1} H_n \left( \frac{1}{n+1} - \frac{1}{n+3} + \frac{1}{n+5} - \ldots \right)$
$$\sum_{n=1}^{\infty} (-1)^{n+1} H_n \left( \frac{1}{n+1} - \frac{1}{n+3} + \frac{1}{n+5} - \ldots \right) = \frac{\pi}{16} \cdot \log(2) + \frac{3}{16} \cdot \log(2) - \frac{\pi^2}{192}$$
$$\sum_{k=...
1
vote
2
answers
64
views
Closed form for $f_k(y)$
The question is quite simple. Given
$$\sum_{k=0}^{n-1}(x+y)^k$$
We can re-write it in terms as a polynomial in $x$, with coefficients being polynomials in $y$, i.e
$$\sum_{k=0}^{n-1}(x+y)^k = \sum_{k=...
- 702
1
vote
3
answers
66
views
I want to use integration for performing summation in Algebra
I am a class 9th student. Sorry if my problem is silly.
I am trying to find the sum of squares from 1 to 10. For this I tried summation, and it was fine.
But now I came to know that Integration can be ...
3
votes
1
answer
143
views
Show that $\sum_{n=1}^{\infty} \frac{(-1)^n (\psi(n) - \psi(2n))}{n} = \frac{\pi^2}{16} + \left(\frac{\ln(2)}{2}\right)^2$
Show That
$$\sum_{n=1}^{\infty} \frac{(-1)^n (\psi(n) - \psi(2n))}{n} =\bbox[15px, #B3E0FF, border: 5px groove #0066CC]{\frac{\pi^2}{16} + \left(\frac{\ln(2)}{2}\right)^2}$$
my work
$$\sum_{n=1}^{\...
3
votes
0
answers
63
views
Is there any function in which the Maclaurin series evaluates to having prime numbered powers and factorials? [duplicate]
I am searching for any information or analysis regarding the functions
$$f(x)=\sum_{n=1}^{\infty}\frac{x^{p\left(n\right)}}{\left(p\left(n\right)\right)!}$$
or
$$g(x)=\sum_{n=1}^{\infty}\frac{\left(-1\...
- 41
5
votes
3
answers
193
views
Show that $\sum_ {k=1}^{\infty}\dfrac{\zeta(2k)-\zeta(3k)}{k}=\ln\left(\frac{2\cosh\left(\sqrt{3}\pi/2\right)}{3\pi}\right)$
Question
$$\zeta(k)=1+\dfrac{1}{2^k}+\dfrac{1}{3^k}+\cdots+\dfrac{1}{n^k}+\cdots$$ Prove that : $$\sum_ {k=1}^{\infty}\dfrac{\zeta(2k)-\zeta(3k)}{k}=\ln\left(\frac{2\cosh\left(\sqrt{3}\pi/2\right)}{...
user1258226
10
votes
3
answers
190
views
Show that $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \frac{1}{n}=\frac{5\zeta(3)}{8}$
$$\sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} \sum_{n=1}^k \dfrac{1}{n}=\frac{5\zeta(3)}{8}$$
I tried to create a proof from some lemmas some are suggested by my Senior friends
Lemma 1 $$
{H_n} = \sum\...
user1254447
2
votes
0
answers
82
views
Limit : $\lim_{n\to+\infty}a^n(n-\zeta(2)-\zeta(3)-\cdots-\zeta(n))$
question
Compute the limit $$\displaystyle{\lim_{n\to+\infty}a^n(n-\zeta(2)-\zeta(3)-\cdots-\zeta(n))}$$, if any, for the various values of the positive real a, where $\zeta$ the zeta function of Mr. ...
0
votes
1
answer
75
views
where is the mistake in my calculations of $\displaystyle \lim_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= \lim_{n\to \infty}a_n$
if $\lim\limits_{n \to \infty}a_n =a$ prove that $\displaystyle \lim_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= a$
define $b_{n-1}= a_n - a_{n-1}$ then $\lim\limits_{n \to \...
- 6,563
1
vote
0
answers
137
views
Simple algebra in rearring terms
I have a very simple mathematical question, and it is just about algebra which seems very tedious. First, let me state my problem from the beginning:
Let $i$ be an index representing countries ($i = {...
- 105
6
votes
2
answers
186
views
Calculate $\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $
question:
how do we find that:
$$ S = \sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $$
I modified the sum
$$\sum\...
user1254447