All Questions
1,809
questions
2
votes
0
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67
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Closed form for $\psi^{1/k}(1)$, where $k$ is an integer
I have proven the identity
$$
\sum_{k=1}^{\infty} \dfrac{\operatorname{_2F_1}(1, 2, 2-1/t,-1/k)}{{k}^{2}} = Γ(2-\dfrac{1}t){\psi^{1/t}(1)}+\psi(-\dfrac{1}t)(\dfrac{1}t(1-\dfrac{1}t))+\gamma(1-\dfrac{1}...
0
votes
0
answers
104
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Is this divergent series, convergent?
Examining the series $\sum_{n=1}^{\infty} \frac{1}{nx}$ alongside its integral counterpart reveals insights into its convergence. Notably, the integral over intervals from $10^n$ to $10^{n+1}$ yields ...
- 81
1
vote
2
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82
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Upper rectangle area sum to approximate 1/x between $1\leq x\leq 3$
I am trying to figure out how to use rectangles to approximate the area under the curve $1/x$ on the interval $[1,3]$ using $n$ rectangle that covers the region under the curve as such.
Here is what I ...
- 1,071
2
votes
2
answers
99
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Prove $\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$
Desmos suggests that$$\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$$Where $\psi$ is the digamma function. I can write the LHS as $$\...
- 6,539
0
votes
0
answers
31
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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
93
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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]
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \log(\tan(\frac{\pi}{4} + x)))}{\tan(2x)} \,dx$$
$$\int_{0}^{\frac{\pi}{4}} \frac{\log(\log(\tan(\frac{\pi}{4} + x))) \cdot \...
2
votes
0
answers
364
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A sum of two curious alternating binoharmonic series
Happy New Year 2024 Romania!
Here is a question proposed by Cornel Ioan Valean,
$$\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
155
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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
$$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) ...
1
vote
1
answer
131
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for any positive numbers $p_k$ how to find the minimum of $\sum_{k=1}^n a_k^2 +(\sum_{k=1}^n a_k)^2$ when $\sum_{k=1}^n p_ka_k=1$? [duplicate]
For any positive numbers $p_k$ how to find the minimum of $\sum\limits_{k=1}^n a_k^2 +(\sum\limits_{k=1}^n a_k)^2$ when $\sum\limits_{k=1}^n p_ka_k=1$?
I saw this problem on my problem book and I ...
- 6,350
1
vote
1
answer
128
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Prove $\sum_{k=j}^{\lfloor n/2\rfloor}\frac1{4^k}\binom{n}{2k}\binom{k}{j}\binom{2k}{k}=\frac1{2^n}\binom{2n-2j}{n-j}\binom{n-j}j$
let $x>1$, $n\in \mathbb N$ and
$$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
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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 $n$ ...
- 6,539
0
votes
1
answer
142
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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 ...
- 6,350
22
votes
3
answers
1k
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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 $m \in \...
- 6,350
-1
votes
1
answer
81
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$\sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{(k\pi )^{4n-1}}}$ [duplicate]
Show that
$${\displaystyle \sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{(k\pi )^{4n-1}}}=\sum _{k=0}^{2n}(-1)^{k-1}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (4n-2k)}{\pi ^{4n-2k}}}\qquad n\in \...
user1235604
1
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2
answers
205
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Which closed form expression for this series involving Catalan numbers : $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$
Obtain a closed-form for the series: $$\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 $\...