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Use the expansion $$\zeta(s)=\frac{1}{s-1}+\gamma+{\cal O}(s-1),$$ hence, since $\zeta(x)=1+2^{-x}+3^{-x}+4^{-x}+5^{-x}+\cdots$, you have $$\zeta(\zeta(x))=\frac{2^x}{1+(2/3)^x+(2/4)^x+(2/5)^x+\cdots} …

You can reduce the evaluation of $S_n$ to a quadrature by means of the Abel-Plana formula, $$S_n=\sum _{s=1}^{n-1} g(s)=\int_1^{n-1}g(s)\,ds+\tfrac{1}{2}g(1)+\tfrac{1}{2}g(n-1)$$ $$\qquad\qquad -\,2\o …

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left …

You might try to regularize the sum, $$S(\alpha)=\sum_{n=1}^\infty \frac{ \cos(\alpha n) \sin(n+1) }{n}= -\tfrac{1}{4} i \left[e^{-i} \ln \left(1-e^{i (\alpha-1)}\right)+e^{-i} \ln \left(1-e^{-i (\alp …

The Gaussian tends to a delta function in the $\sigma\rightarrow 0$ limit, $$ \lim_{\sigma \to 0} \frac{1}{\sqrt{2\pi\sigma^2} } e^{-\frac{1}{2\sigma^2} (xy - \mu)^2} =\delta(xy-\mu)=|y|^{-1}\delta(x- …

The first two moments of $X_i$ are given, $\mathbb{E}[X_i]=\mu_i$, $\mathbb{E}[X_i^2]=\sigma_i^2+\mu_i^2$. We also need the fourth moment, $\mathbb{E}[X_i^4]=\tau_i^4$. Then with $Y =\sqrt{\sum_{i=1}^ …

For any $\delta,x$ such that $0<\delta\leq |x|<\pi$ one has $|F_N(x)|\leq[2\pi(N+1)\sin^2(\delta/2)]^{-1}$, so the error in the delta-function approximation is of order $1/N$. Two explicit examples, i …

The sum $\sum _{n=0}^{\infty}t^{n^2}$ evaluates for $t<1$ to an elliptic theta function, and then taking the limit $t\rightarrow 1$ from below gives $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t …

For each $k>0$, $c\in\mathbb{C}$ it holds that $$\lim_{\alpha\rightarrow\infty}\frac{\Gamma(\alpha+c)}{\Gamma(\alpha)\alpha^c}=1\Rightarrow\lim_{\alpha\rightarrow\infty} \left(\log\frac{\Gamma(k\alpha …

The probability distribution density function of $Y=\cos X$ is $P(y)=(1/\pi)(1-y^2)^{-1/2}$, for $|y|<1$, with moment generating characteristic function $F(s)=J_0(s)$, so the moment generating charact …

Mathematica can actually solve the recursion relation in closed form, $$F_n(n)=-\tfrac{1}{2}(n^2-1)^{-1}\left(\frac{n-1}{n}\right)^n\left[n \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},1,n+1\ …

For a given matrix $V$ this number $Z=p^{-1} \mathbf{1}^\top \mathbf{V}_r \mathbf{V}_r^\top \mathbf{1}$ will converge to $1$ if you take the limit $p\rightarrow\infty$ at fixed $r$. If you average ove …

The OP asks where this recursion relation might appear in a research context. It appears as a discretization of the Emden–Fowler nonlinear differential equation, $$f''(t)=t^{p}[f(t)]^q,$$ for $p=1$, $ …

Q: Is $\lim_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi} $ ? A: use that $r(x)=h(x)+2a(x)$, hence $$\frac{h(x)}{r(x)} = \frac{h(x)/a(x)}{2+h(x)/a(x)}$$ and $$\lim_{x\rightarrow\infty}\frac{h(x) …

I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia, $$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\f …