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6 votes

Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$

This is perhaps overkill but I think the maths is cute and it doesn't require any appeals to Stirling's approximation etc: For $x = 4$, $\displaystyle\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}2^{...
Adam Dougall's user avatar
5 votes

What is the limit of the alternating series $F(z)=\sum_{n=1}^\infty(-1)^n z^{T_n}$ as $z\to1$ for a sequence $T_n\sim cn$?

The answer is no. For example, $$ \sum_{n=1}^\infty (-1)^n z^{4n+(-1)^n} = - \sum_{m=1}^\infty z^{4(2m-1)-1} + \sum_{m=1}^\infty z^{4(2m)+1} = -\frac{z^3}{1-z^8} + \frac{z^9}{1-z^8} = -\frac{z^3+z^5+z^...
Greg Martin's user avatar
  • 81.9k
3 votes

Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$

Stirling's approximation (good for intuition, even if it's not acceptable for an explanation) gives you $$ \frac{4^n(n!)^2}{(2n)!} \approx \frac{4^n(\sqrt{2\pi n} n^n e^{-n})^2} {\sqrt{2\pi(2n)} (...
Matthew Leingang's user avatar
3 votes

Finding the integer part of a sum

The function $n^{2/3}$, with an exponent less than $1$, is everwhere concave downwards for positive $n$: $2n^{2/3}>(n+a)^{2/3}+(n-a)^{2/3},0<a<n$ We also get from AM-GM: $n^{2/3}>[(n+a)(n-...
Oscar Lanzi's user avatar
  • 41.2k
1 vote

The convergence of this series: $\sum\limits_{n=2}^\infty {1\over n^{\log n}}$

Just for your curioisity. It does converge by the integral test $$\int \frac {dx}{x^{\log(x)}}=\int e^{-t^2+t}\,dt=e ^{1/4}\int e^{-\left(t-\frac{1}{2}\right)^2}\,dt$$ $$\int_2^\infty \frac {dx}{x^{\...
Claude Leibovici's user avatar
1 vote

Convergence of $\sum\limits_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}x^n$

We have $$4^n=(1+1)^{2n}>{2n\choose n}={(2n)!\over ( n!)^2}$$ Hence for $ x=\pm 4$ the absolute value of the general term is greater than $1,$ which excludes the convergence. Remark 1 The radius ...
Ryszard Szwarc's user avatar

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