All Questions
1,809
questions
2
votes
0
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67
views
Closed form for $\psi^{1/k}(1)$, where $k$ is an integer
I have proven the identity
$$
\sum_{k=1}^{\infty} \dfrac{\operatorname{_2F_1}(1, 2, 2-1/t,-1/k)}{{k}^{2}} = Γ(2-\dfrac{1}t){\psi^{1/t}(1)}+\psi(-\dfrac{1}t)(\dfrac{1}t(1-\dfrac{1}t))+\gamma(1-\dfrac{1}...
0
votes
0
answers
104
views
Is this divergent series, convergent?
Examining the series $\sum_{n=1}^{\infty} \frac{1}{nx}$ alongside its integral counterpart reveals insights into its convergence. Notably, the integral over intervals from $10^n$ to $10^{n+1}$ yields ...
- 81
1
vote
2
answers
82
views
Upper rectangle area sum to approximate 1/x between $1\leq x\leq 3$
I am trying to figure out how to use rectangles to approximate the area under the curve $1/x$ on the interval $[1,3]$ using $n$ rectangle that covers the region under the curve as such.
Here is what I ...
- 1,071
2
votes
2
answers
99
views
Prove $\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$
Desmos suggests that$$\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$$Where $\psi$ is the digamma function. I can write the LHS as $$\...
- 6,549
0
votes
0
answers
31
views
Can we compare the arithmetic mean of the ratios given the comparison between individual arithmetic means?
I have positive real random numbers $u_1,\ldots,u_n$ and $v_1,\ldots,v_n$ and $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$.
I know that the arithmetic mean of $u_i$'s is greater than the arithmetic mean of $...
- 452
1
vote
1
answer
95
views
Evaluation of $\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$ [closed]
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$$
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \...
2
votes
0
answers
365
views
A sum of two curious alternating binoharmonic series
Happy New Year 2024 Romania!
Here is a question proposed by Cornel Ioan Valean,
$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{2^{2n}}\binom{2n}{n}\sum_{k=1}^n (-1)^{k-1}\frac{H_k}{k}-\sum_{n=1}^{\infty}(-1)...
- 5,495
5
votes
2
answers
157
views
Show that $\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$
Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$$
My try :
We know that
$$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) ...
1
vote
1
answer
131
views
for any positive numbers $p_k$ how to find the minimum of $\sum_{k=1}^n a_k^2 +(\sum_{k=1}^n a_k)^2$ when $\sum_{k=1}^n p_ka_k=1$? [duplicate]
For any positive numbers $p_k$ how to find the minimum of $\sum\limits_{k=1}^n a_k^2 +(\sum\limits_{k=1}^n a_k)^2$ when $\sum\limits_{k=1}^n p_ka_k=1$?
I saw this problem on my problem book and I ...
- 6,620
1
vote
1
answer
129
views
Prove $\sum_{k=j}^{\lfloor n/2\rfloor}\frac1{4^k}\binom{n}{2k}\binom{k}{j}\binom{2k}{k}=\frac1{2^n}\binom{2n-2j}{n-j}\binom{n-j}j$
let $x>1$, $n\in \mathbb N$ and
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}(x+\sqrt{x^2-1}\cos{t})^ndt.$$
Prove that
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{\left(x-\sqrt{x^2-1}\cos{t}\right)^{...
- 137
0
votes
2
answers
80
views
On the convergence of $\sum_{n=0}^\infty(f(x)-T_n\{f\}(x))$
To test the efficiency of the Taylor Series at approximating functions, I was wondering whether$$\sum_{n=0}^\infty(f(x)-T_n\{f\}(x))\tag{$\star$}$$converges, where $T_n\{f\}(x)$ is the degree $n$ ...
- 6,549
0
votes
1
answer
142
views
How to rigorously prove that $e^x = \sum\limits_{n=0}^ \infty \frac{x^n}{n!}$ without defining derivatives? [duplicate]
In my problem book, there was a question: By defining $e= \lim\limits_{n \to \infty}\left( 1+\frac{1}{n} \right) ^n$ prove that $e^x = \sum\limits_{n=0}^ \infty \frac{x^n}{n!}$. this is a strange ...
- 6,620
22
votes
3
answers
1k
views
Conjecture: $\sum\limits_{k=1}^nk^m=S_3(n)\times\frac{P_{m-3}(n)}{N_m}$ for odd $m>1 \ ;\ =S_2(n)\times\frac{P_{m-2}'(n)}{N_m}$ for even $m$.
When I was in high school, I was fascinated by $\displaystyle\sum\limits_{k=1}^n k= \frac{n(n+1)}{2}$ so I tried to find the general solution for $\displaystyle\sum\limits_{k=1}^n k^m$ s.t $m \in \...
- 6,620
-1
votes
1
answer
81
views
$\sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{(k\pi )^{4n-1}}}$ [duplicate]
Show that
$${\displaystyle \sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{(k\pi )^{4n-1}}}=\sum _{k=0}^{2n}(-1)^{k-1}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (4n-2k)}{\pi ^{4n-2k}}}\qquad n\in \...
user1235604
1
vote
2
answers
205
views
Which closed form expression for this series involving Catalan numbers : $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$
Obtain a closed-form for the series: $$\mathcal{S}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$$
From here https://en.wikipedia.org/wiki/List_of_m ... cal_series we know that for $\...
6
votes
4
answers
241
views
Evaluate $\int_{0}^{1}\{1/x\}^2\,dx$
Evaluate
$$\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))}$$, if any, for the various values of the positive real a, where $\zeta$ the zeta function of Mr. ...
0
votes
1
answer
75
views
where is the mistake in my calculations of $\displaystyle \lim_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= \lim_{n\to \infty}a_n$
if $\lim\limits_{n \to \infty}a_n =a$ prove that $\displaystyle \lim_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= a$
define $b_{n-1}= a_n - a_{n-1}$ then $\lim\limits_{n \to \...
- 6,620
1
vote
0
answers
137
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Simple algebra in rearring terms
I have a very simple mathematical question, and it is just about algebra which seems very tedious. First, let me state my problem from the beginning:
Let $i$ be an index representing countries ($i = {...
- 105
6
votes
2
answers
185
views
Calculate $\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $
question:
how do we find that:
$$ 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