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Write $\int_0^1=\int_0^{1/2}+\int_{1/2}^1$ and use $$\int_0^{1/2}\frac{dr}{r\log^n(1+1/r)}<\int_0^{1/2}\frac{dr}{r\log^n(1/r)}=\frac{\log^{1-n}2}{n-1}$$ (the remaining integral is clearly finite).

Let $b_n=\displaystyle\sum_{k=1}^n k^\alpha$ and $c_n=\displaystyle\sum_{k=1}^n a_k$. Then (the "Riemann sum" argument) $$\lim_{n\to\infty}\frac{b_n}{n^{1+\alpha}}=\lim_{n\to\infty}\frac1n\sum_{k=1}^n …

Hint: $$(x-a)^{-k}(x-b)^{-l}=\sum_{j=1}^{k}A_j(x-a)^{-j}+\sum_{j=1}^{l}B_j(x-b)^{-j},$$ where the "$A_j$" part can be obtained from the power series of $(x-b)^{-l}$ around $x=a$: $$(x-b)^{-l}=(a-b)^{- …

$$ {\rm W}(iz)={\rm e}^{z^2}\frac{2}{\sqrt\pi}\int_z^\infty{\rm e}^{-t^2}\,{\rm d}t\underset{t=z+x}{\phantom{\big[}=\phantom{\big]}}\frac{2}{\sqrt\pi}\int_0^\infty{\rm e}^{-x^2-2zx}\,{\rm d}x, $$ $$ \ …

The limit doesn't exist (i.e. it is infinite). For $1\leqslant k\leqslant m_n:=\lfloor\sqrt{n/8}\rfloor$ (and $n$ large enough), let $$i_k:=\lfloor(2k-1)\sqrt{2n}\rfloor=(2k-1)\sqrt{2n}-\epsilon_k.$$ …

My "favourite" counterexample works again. Consider $a_n=(-1)^k$ for $2^k\leqslant n<2^{k+1}$, $k\geqslant 0$. Then, for $f(x):=\sum_{n=1}^\infty a_n x^n$, we get $g(x):=(1-x)f(x)=x+2\sum_{k=1}^\infty …

Just an idea (not a complete answer) is to use $\Im\sqrt{x+iy}=\sqrt{(\sqrt{x^2+y^2}-x)/2}$ for real $x,y\geqslant 0$ (and the principal branch of the square root on the left). So the substitution $x= …

I'm not completely sure what the integration domain is. If it is really supposed that all variables are ranging from $0$ to $1$, then $1-x_1-\ldots-x_n$ gets negative when all of $x_1,\ldots,x_n$ are …

Mixed indices confuse. $\displaystyle\frac{\partial}{\partial H_{uv}}\sum_{i,j}\log\frac{\sum_k W_{ik} H_{kj}}{\text{const.}} = \sum_i\frac{W_{iu}}{\sum_k W_{ik} H_{kv}}$.

(A proof, not very "natural" though.) $f(x)=1-\cos^{2/3}x$ satisfies $xf''(x)=g(x)f'(x)$, where $$g(x)=x\cot x+\frac{x}3\tan x=1+\sum_{n=1}^\infty g_n x^n$$ with $g_n=0$ for odd $n$, and $3g_{2n}=(-1) …

Use Euler–Maclaurin summation formula with remainder. For any $\beta>0$ \begin{align} \sum_{k=1}^n\log(\beta+k/n)&=\int_0^n\log(\beta+x/n)\,dx \\&+\frac{\log(\beta+1)-\log\beta}2-\frac1{12n\beta(\beta …

I doubt there's a simple closed form. What I get (easily) is a series using Bessel functions. Taking the following known generating function $$\sum_{n\in\mathbb{Z}}J_n(z)t^n=\exp\left[\frac{z}{2}\left …

Suppose $\sum_{n=1}^\infty|a_n|$ converges. Then $\left(\sum_{n=1}^\infty a_n\right)^2=\sum_{n=1}^\infty a_n^2+2\sum_{n=1}^\infty a_n\sum_{k=1}^\infty a_{n+k}$. Replacing $a_n$ with $(-1)^n a_n$ and r …

Another not-an-answer regarding the computational matter. As commented, $$I=\int_0^\pi\frac{\sqrt{\sin x}}{x}\,dx=\int_0^\pi\frac{\sqrt{\sin x}}{\pi-x}\,dx =\frac\pi2\int_0^\pi\frac{\sqrt{\sin x}}{x(\ …

\begin{align*} 2u_k(x)\sin\frac{x}2&=k\sin\frac{x}2+\sum_{n=1}^{k-1}(k-n)\left[\sin\left(n+\frac12\right)x-\sin\left(n-\frac12\right)x\right] \\&=\sum_{n=\color{red}{0}}^{k-1}(k-n)\sin\left(n+\frac12\ …