New posts in Sequences And Series

Is there a closed form for $\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}?$

Let $P(x)=a_0+a_1x+a_2 x^2+a_3x^3+.......+a_nx^n$ and $P(1)=4$ and $P(5)=136$

Computing $\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$

An interesting limit: $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}$

How do you prove $\pi =\sqrt{12}\sum_{n\ge 0}\frac{(-1)^n}{3^n(2n+1)}$?

Iterative algorithm for $\pi$?

Show $1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$.

Calculate: $\int_0^\infty [x]e^{-x} \, dx$ where $[x]:=\max \{k\in\mathbb{Z}:k\leq x\}$

Prove: $\int_0^1 \int_0^1 \frac{\ln{\left(2+yx\right)}}{1+yx} \; \mathrm{d}y\; \mathrm{d}x = \frac{13}{24} \zeta(3)$

How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$