Summation problem (associated with use of Stirling's Approx)

Question

Prove that $$\sum_{x=0}^∞ \frac{(2x)!}{({2^{2x-2}})(1-4x^2)(x!)^2} = \pi$$

So I have solved this using Stirling's approach:

In general $$x!\approx\sqrt {(2\pi x)}(\frac {{x}^x}{{e}^x})$$

So we can eliminate factorials of $2x!$ and $x!$, leaving only one $x!$ at denominator

$$4\sqrt 2\sum_{x=0}^∞ \frac{{x}^x}{{e}^x(1-4x^2)(x!)}= S_1$$

We know that $${e}^{-x}=\sum_{n=0}^∞ \frac {({-1})^n{x}^n}{n!}=S_2$$

And I attempt to combine this sum above $S_2$ into $S_1$, but it turns out to be a very complicated expression with two almost identical sums $$S_1=4\sqrt 2\sum_{x=0}^∞ \frac{{x}^x{\sum_{n=0}^∞ \frac {({-1})^n{x}^n}{n!}}}{(1-4x^2)(x!)}$$

These two almost identical sums I am referring to are :

$${e}^{-x}=\sum_{n=0}^∞ \frac {({-1})^n{x}^n}{n!}=S_2$$ and $$\sum_{x=0}^∞ \frac {{x}^x}{x!}=S_3$$

Can you guys suggest the next step I should do to simplify this problem?

Answer 1

Your approach will not work, as, in general $a_n\approx b_n$ for some pair of sequences, $\left(a_n\right)_{n\in \mathbb Z^+}$ and $\left(b_n\right)_{n\in \mathbb Z^+}$, does not imply $\sum_{i=1}^{\infty}a_n=\sum_{i=1}^{\infty}b_n$.

For instance, take $a_n=\frac{1}{(n+1)^2-1}$ and $b_n=\frac{1}{n^2}$ for all positive integer values of $n$. Since $\lim_{n\rightarrow\infty}{\frac{(n+1)^2-1}{n^2}}=1$, following your use of the Stirling approximation for this problem, it would be reasonable to say that $a_n\approx b_n$. But, it is easy to show that $\sum_{i=1}^{\infty}a_n=1$ while $\sum_{i=1}^{\infty}b_n$ famously equals $\frac{\pi^2}{6}$.

However, there is a direct way to easily solve the problem.

I'll first rewrite the series using more traditional indices: $$\sum_{n=0}^{\infty}\frac{(2n)!}{\left(2^{2n-2}\right)\left(1-4n^2\right)\left(n!\right)^2}$$

Now, if you use the fact that $1-4n^2=(1-2n)(1+2n)$, we can apply a partial fraction decomposition on the series:

$$=\sum_{n=0}^{\infty}\frac{(2n)!}{\left(2^{2n-1}\right)\left(2n+1\right)\left(n!\right)^2}-\sum_{n=0}^{\infty}\frac{(2n)!}{\left(2^{2n-1}\right)\left(2n-1\right)\left(n!\right)^2}$$

It is worth noting that for us to be able to actually split the series up into two like we have done above, each part needs to converge. While we don't know whether or not they do at this point, we'll push forth anyway (you can avoid this point entirely if you want by continuing to use a single sum throughout the proof - the changes are minor. I have just broken it up like this for simplicity.)

At this point, we may use the fact that, for any real $k>0$: $$\frac{1}{k}=\int_0^1 x^{k-1} dx$$

If we use this fact on the two series we have obtained, being careful with the $0^{th}$ index term of the second series, we can manipulate them into:

$$=\sum_{n=0}^{\infty}\frac{(2n)!\int_0^1 x^{2n} dx}{\left(2^{2n-1}\right)\left(n!\right)^2}-\sum_{n=1}^{\infty}\frac{(2n)!\int_0^1 x^{2n-2} dx}{\left(2^{2n-1}\right)\left(n!\right)^2}+2$$$$=\int_0^1\left(\sum_{n=0}^{\infty}\frac{(2n)!x^{2n}}{\left(2^{2n-1}\right)\left(n!\right)^2}\right)dx-\int_0^1\left(\sum_{n=1}^{\infty}\frac{(2n)!x^{2n-2}}{\left(2^{2n-1}\right)\left(n!\right)^2}\right)dx+2$$$$=2\int_0^1\left(\sum_{n=0}^{\infty}\frac{(2n)!\left(\frac{x^2}{4}\right)^n}{\left(n!\right)^2}\right)dx-2\int_0^1\frac{1}{x^2}\left(\sum_{n=1}^{\infty}\frac{(2n)!\left(\frac{x^2}{4}\right)^n}{\left(n!\right)^2}\right)dx+2$$$$=2\int_0^1\frac{x^2-1}{x^2}\left(\sum_{n=1}^{\infty}\frac{(2n)!\left(\frac{x^2}{4}\right)^n}{\left(n!\right)^2}\right)dx+4$$

Now, recognizing that $\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$, we get: $$=2\int_0^1\frac{x^2-1}{x^2}\left(\sum_{n=1}^{\infty}\binom{2n}{n}\left(\frac{x^2}{4}\right)^n\right)dx+4$$

Then, we may finally use the formula here for the generating function of the central binomial coefficient to reduce the series too: $$=2\int_0^1\frac{x^2-1}{x^2}\left(\frac{1}{\sqrt{1-4\left(\frac{x^2}{4}\right)}}-1\right)dx+4$$$$=2\int_0^1\frac{1-x^2-\sqrt{1-x^2}}{x^2}dx+4$$

Now, if we perform the substitution $x=\operatorname{sin}(\theta)$, then $dx=\operatorname{cos}(\theta)d\theta$, so we get: $$=2\int_0^{\frac{\pi}{2}} (\operatorname{cos}(\theta)-1)\operatorname{cot}^2(\theta)+4$$

One can further simplify, but using Wolfram alpha is easier (the problem is merely one of integration now): $$=2\left[\theta-\operatorname{sin}(\theta)-\frac{1}{2}\operatorname{tan}(\frac{\theta}{2})-\frac{1}{2}\operatorname{cot}(\frac{\theta}{2})+\operatorname{cot}(\theta)\right]_0^{\frac{\pi}{2}}+4$$ which simplifies to plain old $$=\pi$$

Answer 2

Another possible way .
Let's use $k$ in place of $x$ and write $$ \eqalign{ & S = \sum\limits_{0\; \le \;k} {{{\left( {2k} \right)!} \over {2^{\,2k - 2} \left( {1 - 4k^{\,2} } \right)\left( {k!} \right)^{\,2} }}} = \cr & = \sum\limits_{0\; \le \;k} {{{\Gamma \left( {2k + 1} \right)} \over {2^{\,2k - 2} \left( {1 - 4k^{\,2} } \right) \Gamma \left( {k + 1} \right)\Gamma \left( {k + 1} \right)}}} \cr} $$

Then we can use the Duplication formula for the Gamma function $$ \Gamma \left( {2\,z + 1} \right) = {{2^{\,2\,z} } \over {\sqrt \pi }}\Gamma \left( {z + 1/2} \right)\Gamma \left( {z + 1} \right) $$ to get $$ \eqalign{ & S = \sum\limits_{0\; \le \;k} {{{\Gamma \left( {2k + 1} \right)} \over {2^{\,2k - 2} \left( {1 - 4k^{\,2} } \right) \Gamma \left( {k + 1} \right)\Gamma \left( {k + 1} \right)}}} = \cr & = {1 \over {\sqrt \pi }}\sum\limits_{0\; \le \;k} {{{2^{\,2\,k} \Gamma \left( {k + 1/2} \right) \Gamma \left( {k + 1} \right)} \over {2^{\,2k - 2} \left( {1 - 4k^{\,2} } \right)\Gamma \left( {k + 1} \right)\Gamma \left( {k + 1} \right)}}} = \cr & = - {1 \over {\sqrt \pi }}\sum\limits_{0\; \le \;k} {{{\Gamma \left( {k + 1/2} \right)} \over {\left( {k + 1/2} \right)\left( {k - 1/2} \right)\Gamma \left( {k + 1} \right)}}} = \cr & = - {1 \over {\sqrt \pi }}\sum\limits_{0\; \le \;k} {{{\Gamma \left( {k - 1/2} \right)} \over {\left( {k + 1/2} \right)\Gamma \left( {k + 1} \right)}}} = \cr & = - {1 \over {\sqrt \pi }}\sum\limits_{0\; \le \;k} {{{\Gamma \left( {k - 1/2} \right)} \over {{{\Gamma \left( {k + 3/2} \right)} \over {\Gamma \left( {k + 1/2} \right)}}\Gamma \left( {k + 1} \right)}}} = \cr & = - {1 \over {\sqrt \pi }}\sum\limits_{0\; \le \;k} {{{\Gamma \left( {k + 1/2} \right)\Gamma \left( {k - 1/2} \right)} \over {\Gamma \left( {k + 3/2} \right)}}{1 \over {k!}}} = \cr & = - {1 \over {\sqrt \pi }}{{\Gamma \left( {1/2} \right)\Gamma \left( { - 1/2} \right)} \over {\Gamma \left( {3/2} \right)}}\;{}_2F_{\,1} \left( {\left. {\matrix{ {1/2,\; - 1/2} \cr {3/2} \cr } \;} \right|\;1} \right) = \cr & = 4{{\Gamma (3/2)\Gamma (3/2)} \over {\Gamma (1)\Gamma (2)}} = \pi \cr} $$ where in the last step we have applied to the Hypergeometric function the Gauss theorem.

Answer 3

We know that $$\frac{d^n}{dx^n}(1-x)^{-1/2}=\frac{(2n)!}{2^{2n}n!}(1-x)^{-n-1/2}$$ Because it's true for $n=0$ and if true for any $n\ge0$ then $$\begin{align}\frac{d^{n+1}}{dx^{n+1}}(1-x)^{-1/2}&=\frac d{dx}\frac{(2n)!}{2^{2n}n!}(1-x)^{-n-1/2}\\ &=\frac{(2n)!}{2^{2n}n!}(-(n+1/2))(-1)(1-x)^{-n-3/2}\\ &=\frac{(2n)!}{2^{2n}n!}\frac{(2n+2)(2n+1)}{2(n+1)(2)}(1-x)^{-n-3/2}\\ &=\frac{(2(n+1))!}{2^{2(n+1)}(n+1)!}(1-x)^{-(n+1)-1/2}\end{align}$$ So true for $n+1$. So the Taylor series is $$(1-x)^{-1/2}=\sum_{n=0}^{\infty}\left.\frac{d^n}{dx^n}(1-x)^{-1/2}\right|_{x=0}\frac{(x-0)^n}{n!}=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^2}x^n$$ That's so close to the target series that pretty much all we have to do is multiply by $x^{-3/2}$ and integrate twice and we'll be there. So $$\int_0^tx^{-3/2}\left(1-(1-x)^{-1/2}\right)dx=\frac{2(1-t)^{1/2}-2}{t^{1/2}}=-\sum_{n=1}^{\infty}\frac{(2n)!}{2^{2n}(n-1/2)(n!)^2}t^{n-1/2}$$ As can be verified by differentiation and checking that $$\lim_{x\rightarrow0^+}\frac{2(1-x)^{1/2}-2}{x^{1/2}}=0$$ by rationalizing the numerator. Then if we let $t=\sin^2\theta$ we have $$\begin{align}\int_0^1\frac{2(1-t)^{1/2}-2}{t^{1/2}}dt&=\int_0^{\pi/2}(4\cos^2\theta-4\cos\theta)d\theta=\int_0^{\pi/2}(2+2\cos2\theta-4cos\theta)d\theta\\ &=\pi-4=\left.\sum_{n=1}^{\infty}\frac{(2n)!}{2^{2n}(1/2-n)(1/2+n)(n!)^2}t^{n+1/2}\right|_0^1\\ &=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(1/2-n)(1/2+n)(n!)^2}-4\end{align}$$ Where we have restored the omitted $n=0$ term.

Answer 4

Another way to do it $$a_n= {{{\left( {2n} \right)!} \over {2^{\,2n - 2} \left( {1 - 4n^{\,2} } \right)\left( {n!} \right)^{\,2} }}} =-\frac{2 \Gamma \left(n-\frac{1}{2}\right)}{\sqrt{\pi } (2 n+1) \Gamma (n+1)} $$

Now, if you recognize the infinite series, $$S_\infty=\sum_{n=0}^\infty a_n x^{2n}=2 \sqrt{1-x^2}+\frac{2 \sin ^{-1}(x)}{x}$$ Make $x=1$ for the result.

We could even compute the partial sums

$$S_p=\sum_{n=0}^p a_n x^{2n}=\sum_{n=0}^\infty a_n x^{2n}-\sum_{n=p+1}^\infty a_n x^{2n}=S_\infty-\sum_{n=p+1}^\infty a_n x^{2n}$$ $$T_p(x)=\sum_{n=p+1}^\infty a_n x^{2n}=-\frac{x^{2 p+2} \Gamma \left(p+\frac{1}{2}\right) \Gamma \left(p+\frac{3}{2}\right) }{ \sqrt{\pi }}\, _3\tilde{F}_2\left(1,p+\frac{1}{2},p+\frac{3}{2};p+2,p+\frac{5}{2};x^2\right)$$

For $x=1$ we have $T_p=\pi-\frac{a_p}{b_p}$ which give the sequences $$\{10,97,1343,32057,703049,36485783,72872467,19801986719,751927973347\}$$ $$\{3,30,420,10080,221760,11531520,23063040,6273146880,238379581440\}$$ which are not in $OEIS$.