How would one calculate the value of $$ \sum_{k=1}^{\infty}\dfrac{(2k)!}{(k!)^2}, $$ which by the ratio test is easily seen to be convergent? Is there a general strategy for calculating infinite series, other than mathematical maturity?
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4"which by the ratio test is easily seen to be convergent?" When I applied it, I got the conclusion that the series diverges. – projectilemotion Jan 13 '18 at 15:52
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My bad you're absolutely right! – EEEB Jan 13 '18 at 15:52
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$\binom{2k}{k}$ does not tend to zero as $k\to +\infty$, hence the given series is blatantly divergent. – Jack D'Aurizio Jan 13 '18 at 15:54
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Let $\left(a_n\right)_{n \in \mathbb{N}}$ the sequence define by $$ a_n=\frac{\left(2k\right)!}{\left(k!\right)^2} $$ It is stricly positive and $$ \frac{a_{n+1}}{a_n}=\frac{\left(2k+2\right)\left(2k+1\right)\left(2k\right)!}{\left(k+1\right)^2\left(k!\right)^2}\frac{\left(k!\right)^2}{\left(2k\right)!}\underset{(+\infty)}{\sim}4>1 $$ The series easily diverges. However, with $\displaystyle b_n=\frac{1}{a_n}$ $$ \sum_{n=0}^{+\infty}\frac{\left(n!\right)^2}{\left(2n\right)!}=\frac{2}{27}\left(18+\sqrt{3}\pi\right) $$
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This kind of sum is not that easy to calculate. The following proof is really interesting and I suggest you try to prove what I will admit. In fact, you can write that $$ \sum_{n=0}^{+\infty}\frac{\left(n!\right)^2}{\left(2n\right)!}=\sum_{n=0}^{+\infty}\frac{1}{\displaystyle \binom{2n}{n}} $$ If you study the function $f$ defined on $\left]-1,1\right[$ by $$ f\left(x\right)=\frac{\text{arcsin}\left(x\right)}{\sqrt{1-x^2}} $$ You can find a simple differential equation verified by $f$, but you can also search for power series that are also solution of this. By unicity of power series development you will deduce for $x \in \left]-1,1\right[$ $$ f\left(x\right)=\sum_{n=1}^{+\infty}\frac{2^{2n-1}}{\displaystyle n\binom{2n}{n}}x^{2n-1} $$ Then, you can deduce the development on $\left]-1,1\right[$ of $$ g\left(x\right)=\frac{x^2}{1-x^2}+\frac{x\text{arcsin}\left(x\right)}{\left(1-x^2\right)^{3/2}} $$ which is $$ g\left(x\right)=\sum_{n=1}^{+\infty}\frac{\left(2n-1\right)}{\displaystyle \left(n+1\right)}\frac{2^{2n+1}}{\displaystyle \binom{2n+2}{n+1}}x^{2n} $$ For $\displaystyle x=\frac{1}{2}$, $$ g\left(\frac{1}{2}\right)=\sum_{n=1}^{+\infty}\frac{1}{\displaystyle \binom{2n}{n}}=\frac{1}{4}\times\frac{4}{3}+\frac{\pi}{12}\left(\frac{4}{3}\right)^{3/2}=\frac{2}{27}\left(18+\sqrt{3}\pi\right) $$
Atmos
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$$ \sum_{n\geq 0}\binom{2n}{n}^{-1}$$ can be simply tackled through Euler's Beta function, since $\binom{2n}{n}^{-1}=(2n+1) B(n+1,n+1) = (2n+1)\int_{0}^{1}x^n(1-x)^n,dx$ and $\sum_{n\geq 0}(2n+1)z^n =\frac{1+z}{(1-z)^2}$. – Jack D'Aurizio Jan 13 '18 at 16:39
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Here is a list of more approaches (There was a question regarding this sum): Evaluating:$\sum_{n=0}^\infty\frac1{\binom{2n}{n}}$ and Calculate $\sum\limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$ – projectilemotion Jan 13 '18 at 16:41
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I don't think that this series is convergent. Because we have $$\sum_{k=1}^{\infty}\frac{(2k)!}{(k!)^2} = \sum_{k=1}^{\infty}\frac{(2k)\cdot (2k-1)\cdot...\cdot k!}{(k!)^2} = \sum_{k=1}^{\infty}\frac{(2k)\cdot (2k-1)\cdot...\cdot (k+1)}{k!} = \sum_{k=1}^{\infty}\binom{2k}{k}$$ which is obviously divergent.
ArsenBerk
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The only series whose sum we can calculate are Geometric series and maybe telescopic ones? – EEEB Jan 13 '18 at 15:57
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3@EEEB Definitely not. For example: $\sum\limits_{n=0}^{\infty} \dfrac{1}{n!}=e$ is neither geometric, nor telescopic. – projectilemotion Jan 13 '18 at 16:00
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1@projectilemotion is right, in general Taylor Series of certain functions can also be calculated. Except from that, I know that there are methods (like Integral Test) used for approximating the sum of more complicated series. – ArsenBerk Jan 13 '18 at 16:05