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Call a function $F$ nice, if $$\DeclareMathOperator{\Dm}{d\!} \begin{align} &\zeta(2)\sum_{\frac{a}{b}\in\mathbb{Q}_n} \frac{F(\log \frac{a}{b})}{\sqrt{abn}}\xrightarrow[n\to \infty]{}\int F(x)\Dm x …

The same value as in Nemo's answer may be obtained in a slightly more elementary way: denote $c_k={n\choose k} (k+1)^{n-2}$. We want to find the limit of $\frac{\sum (n-1)c_k}{\sum (k+1)c_k}$. Both su …

Yes, you may consider them as a module over the ring of polynomials. In other words, you may sum up such series and multiply them by polynomials. Also, you may differentiate such series, and usual rul …

Yes, the limit exists and equals $1-\gamma$. Summation may be taken by all $p$, it gives the same sum. At first, we estimate the sum over $k>0$. Denote $q=(p-1)p^k$. Estimate from above $\ln p\leqsla …

Yes, $b_n$ must be bounded. Assume the contrary. Passing to a subsequence we may suppose that $a_n\to a$, $b_n\to \infty$. We have $$\lambda_n=\frac{b_n-c_n}{b_n-a_n}\to 1;\, 1-\lambda_n=\frac{c_n-a_n …

Yes, it is $\lambda$. At first, if $n$ is such that $\lambda 2^{-n}<1$ (therefore, for all large enough $n$) we have $(1-\lambda 2^{-n})^{2^k}\geqslant 1-\lambda 2^{k-n}$ be Bernoulli inequality that …

If $\Re z>0$, the series $\sum_n \binom{z}n$ converges absolutely (by Raabe test, for example), thus we may replace $x$ to 0 and just need to check that this sum equals $2^z$. This follows from anothe …

Denote $$A=\sum_{p_1+\ldots+p_k=S} \lambda_1^{|p_1|}\lambda_2^{|p_2|}\ldots \lambda_k^{|p_k|}.\quad (1)$$ We prove two things: 1) Your limit equals $A$. 2) $A$ equals to what you write. Start with …

Note that $$ f(N):=\left(\frac{\Gamma(N+p)}{\Gamma(N)}\right)^{n+1} $$ is a unitary polynomial in $N$ of degree $pn+p$: $f(N)=N^{pn+p}+\dots$. Its $(p-1)$-st finite difference $$ \Delta^{p-1}f(N)=\su …

If the limit exists, it of course equals 1. Indeed, $A(m):=a(m)a(m+1)\dots a(2m-1)=ds_{10}(3^m)/ds_{10}(3^{2m})\in [\frac1{9m},9m]$. But if $\lim a(m)\ne 1$, then $A(m)$ is either exponentially large …

If $B$ can remove an area not more than $c n$, then there exist at most 1 bricks of size between $cn/2$ and $n$, at most 2 bricks of size between $cn/3$ and $cn/2$, and so on. Totally, the whole siz …

Fix $\varepsilon>0$. For any positive integer $n$ consider the (closed) set $E_n$ of $u\in [0,4a]$ such that $|f(x+u)-f(x)|\leq \varepsilon$ for all $x\geq n$. Then $[0,4a]=\cup_n E_n$, thus there exi …

Power means of order $1/n$ converge to the geometric mean when $n$ tends to $+\infty$. You ask about the weighted version of this fact, when we have the numbers $a, b$ with weights $\gamma$, $1-\gamma …

I think, if $c=np+o(n)$, you may simply use $P(X=c+1)=P(X=c)(1+o(1))$, and so $P(X\in \{c,c+1,\dots,c+M-1\})=(M+o(1))P(X=c)$ for any fixed $M$.

Some estimates like $\alpha_n(s)\leqslant A(T)^n/n!$ for all $s\in [0,T]$ should hold by induction, if we suppose something moderate on your functions, which depend on $\sigma$. This yields desired un …