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$$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$$ Beta function

$${1\over {2n\choose n}}=n\int_{0}^{1}x^{n-1}(1-x)^ndx$$

$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}}=\int_{0}^{1}{2(1-t)\over [1-2t(1-t)]^2}dt$$

We can split out the sum

$$\sum_{n=0}^{\infty}{2^{n-1}(n^2-n\pi+1)\over (2n+1){2n\choose n}}-\sum_{n=0}^{\infty}{2^{n-1}(n^2+n-1)\over (2n+3){2n\choose n}}=1$$

With out the denominator $2n+1$ and $2n+3$ We can apply the beta function directly and integrate. I don't how to get the factors $2n+1$ and $2n+3$ with the beta function. Any hints on how to prove S?

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This answer is divided into several steps.

Overview:

At first we show \begin{align*} \sum_{n=1}^\infty\frac{(2x)^{2n}}{n\binom{2n}{n}}=\frac{2x}{\sqrt{1-x^2}}\arcsin(x)\qquad\qquad |x|<1\tag{1} \end{align*}

With a small trick we derive the generating function \begin{align*} \sum_{n=0}^\infty\frac{(2x)^{2n}}{(2n+1)\binom{2n}{n}} =\frac{1}{x\sqrt{1-x^2}}\arcsin(x) \end{align*} just as intermediate step to derive the following series as basis for further calculations: \begin{align*} A(x):=\frac{1}{2}\sum_{n=0}^\infty\frac{(2x)^{n}}{(2n+3)(2n+1)\binom{2n}{n}} =\frac{1}{x}-\frac{1}{x}\sqrt{\frac{2}{x}-1}\cdot\arcsin\left(\sqrt{\frac{x}{2}}\right) \end{align*}

Since we have to additionally respect the factor $(n^2-n\pi+1)(n^2+n-1)$ in the numerator of OPs series we apply the differential operator $D_x$ and calculate from $A(x)$ \begin{align*} (xD_x)^kA(x)&=\frac{1}{2}\sum_{n=0}^\infty\frac{n^k(2x)^{n}}{(2n+3)(2n+1)\binom{2n}{n}}\qquad\qquad k=1,\ldots,4 \end{align*} These are the building blocks to finally calculate OPs series.

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Generating function with reciprocal central binomial coefficient:

The central binomial coefficient can be represented with the Betafunction as \begin{align*} \frac{1}{\binom{2n}{n}}&=\frac{\left(n!\right)^2}{(2n)!}=\frac{n\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}\\ &=n\beta(n+1,n)\\ &=n\int_0^1t^n(1-t)^{n-1} \end{align*}

We obtain \begin{align*} \sum_{n=1}^\infty\frac{(2x)^{2n}}{n\binom{2n}{n}}&=\sum_{n=1}^\infty(2x)^{2n}\int_0^1t^n(1-t)^{n-1}dt\\ &=\frac{1}{1-t}\int_0^1\sum_{n=1}^\infty(4x^2t)^n(1-t)^{n}dt\\ &=\frac{1}{1-t}\int_0^1\frac{4x^2t}{1-4x^2t(1-t)}dt \end{align*}

Substituting $x(2t-1)=s$ gives \begin{align*} x(2t-1)&=s\qquad &x^2(4t^2-4t+1)&=s^2\\ 2xdt&=ds\qquad &4x^2t^2-4x^2t&=s^2-x^2 \end{align*}

We get \begin{align*} \sum_{n=1}^\infty\frac{(2x)^{2n}}{n\binom{2n}{n}}&=\int_{-x}^x\frac{s+x}{s^2+(1-x^2)}ds\\ &=\left.\frac{1}{2}\log(s^2-x^2+1)\right|_{-x}^{x}+\left.\frac{x}{1-x^2}\arctan\left(\frac{s}{\sqrt{1-x^2}}\right)\right|_{-x}^x\\ &=\frac{2x}{\sqrt{1-x^2}}\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)\\ &=\frac{2x}{\sqrt{1-x^2}}\arcsin\left(\frac{\frac{x}{\sqrt{1-x^2}}}{\sqrt{\frac{x^2}{1-x^2}+1}}\right)\\ &=\frac{2x}{\sqrt{1-x^2}}\arcsin\left(x\right)\\ \end{align*} and (1) follows.

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Generating function of $A(x)$:

We derive a generating function for

\begin{align*} A(x)=\frac{1}{2}\sum_{n=0}^\infty\frac{(2x)^{n}}{(2n+3)(2n+1)\binom{2n}{n}}\tag{2} \end{align*}

We do it in three steps. At first we show \begin{align*} \sum_{n=0}^\infty\frac{(2x)^{n}}{(2n+1)\binom{2n}{n}}=\frac{1}{x\sqrt{1-x^2}}\arcsin(x)\tag{3} \end{align*}

Since \begin{align*} \frac{1}{2n+1}\binom{2n}{n}^{-1}&=\frac{1}{2n+1}\cdot\frac{(n!)^2}{(2n)!}\cdot\frac{2(n+1)^2}{2(n+1)^2}\\ &=\frac{2}{n+1}\cdot\frac{((n+1)!)^2}{(2n+2)!}\\ &=\frac{2}{n+1}\binom{2n+2}{n+1}^{-1}\tag{4} \end{align*}

we get from (1) \begin{align*} \sum_{n=0}^\infty\frac{(2x)^{2n}}{(2n+1)\binom{2n}{n}} &=2\sum_{n=0}^\infty\frac{(2x)^{2n}}{(n+1)\binom{2n+2}{n+1}}\\ &=2\sum_{n=1}^\infty\frac{(2x)^{2n-2}}{n\binom{2n}{n}}\\ &=\frac{1}{x\sqrt{1-x^2}}\arcsin(x)\\ &=1+\frac{2}{3}x^2+\frac{8}{15}x^4+\frac{16}{35}x^6+\frac{128}{315}x^8+\cdots \end{align*}

Substituting $n$ with $n+1$ in (4) gives

\begin{align*} \frac{1}{2n+3}\binom{2n+2}{n+1}^{-1}=\frac{2}{n+2}\binom{2n+4}{n+2}^{-1} \end{align*}

We get from (2) \begin{align*} \sum_{n=0}^\infty\frac{(2x)^{2n}}{(2n+3)\binom{2n+2}{n+1}} &=2\sum_{n=0}^\infty\frac{(2x)^{2n}}{(n+2)\binom{2n+4}{n+2}}\tag{5}\\ &=2\sum_{n=2}^\infty\frac{(2x)^{2n-4}}{n\binom{2n}{n}}\\ &=\frac{1}{4x^3\sqrt{1-x^2}}\arcsin(x)-\frac{1}{4x^2}\\ &=\frac{1}{6}+\frac{2}{15}x^2+\frac{4}{35}x^4+\frac{32}{315}x^6+\frac{64}{693}x^8+\cdots \end{align*}

On the other hand we get from (4) \begin{align*} \frac{1}{2n+3}\binom{2n+2}{n+1}^{-1}=\frac{1}{2n+3}\cdot\frac{n+1}{2n+1}\cdot \frac{1}{2}\binom{2n}{n}^{-1} \end{align*}

and from this identity we obtain from (5) \begin{align*} \sum_{n=0}^\infty\frac{(2x)^{2n}}{(2n+3)\binom{2n+2}{n+1}} &=\frac{1}{2}\sum_{n=0}^\infty\frac{(n+1)(2x)^{2n}}{(2n+3)(2n+1)\binom{2n}{n}}\tag{6}\\ &=\frac{1}{4x^3\sqrt{1-x^2}}\arcsin(x)-\frac{1}{4x^2} \end{align*}

We want to get rid of $n+1$ in the numerator in (6) in order to obtain the series $A(x)$. We substitute \begin{align*} x\rightarrow \sqrt{\frac{u}{2}} \end{align*} and then we integrate the series.

We obtain from (6) by this substitution \begin{align*} \sum_{n=0}^\infty\frac{2^{n-1}(n+1)}{(2n+3)(2n+1)\binom{2n}{n}}u^n &=\frac{1}{4\left(\frac{u}{2}\right)^{\frac{3}{2}}\sqrt{1-\frac{u}{2}}}\arctan\left(\sqrt{\frac{u}{2}}\right)-\frac{1}{2u}\tag{7}\\ &=\frac{1}{u\sqrt{u(2-u)}}\arcsin\left(\sqrt{\frac{u}{2}}\right)-\frac{1}{2u}\\ &=\frac{1}{6}+\frac{1}{15}u+\frac{1}{35}u^2+\frac{4}{315}u^3+\frac{4}{693}u^4+\frac{8}{3003}u^5+\cdots \end{align*}

Integrating the RHS of (7) with the help of Wolfram Alpha gives $A(u)$.

\begin{align*} A(u)&=\sum_{n=0}^\infty\frac{2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\ &=\frac{1}{u}-\frac{1}{u}\sqrt{\frac{2}{u}-1}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\tag{8}\\ &=\frac{1}{6}+\frac{1}{30}u+\frac{1}{105}u^2+\frac{1}{315}u^3+\frac{4}{3465}u^4+\frac{4}{9009}u^5+\cdots\\ \end{align*} and the claim (2) follows.

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Building blocks: $(uD_u)^kA(u)$

The next step is to successively apply the operator $uD_u$ on $A(u)$ with $D_u$ the differential operator. We obtain with some help of Wolfram Alpha \begin{align*}\ (uD_u)A(u)&=\sum_{n=0}^\infty\frac{n2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\ &=-\frac{3}{2u}+\frac{3-u}{u^2\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\tag{9}\\ &=\frac{1}{30}u+\frac{2}{105}u^2+\frac{1}{105}u^3+\frac{16}{3465}u^4+\frac{20}{9009}u^5+\cdots\\ \\ (uD_u)^2A(u)&=\sum_{n=0}^\infty\frac{n^22^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\ &=\frac{4u-9}{2u(u-2)}+\frac{u^2-7u+9}{u^2(u-2)\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\tag{10}\\ &=\frac{1}{30}u+\frac{4}{105}u^2+\frac{1}{35}u^3+\frac{64}{3465}u^4+\frac{100}{9009}u^5+\cdots\\ \\ (uD_u)^3A(u)&=\sum_{n=0}^\infty\frac{n^32^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\ &=-\frac{5u^2-25u+27}{2u(u-2)^2}-\frac{u^3-13u^2+34u-27}{u^2(u-2)^2\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\tag{11}\\ &=\frac{1}{30}u+\frac{8}{105}u^2+\frac{3}{35}u^3+\frac{256}{3465}u^4+\frac{500}{9009}u^5+\cdots\\ \\ (uD_u)^4A(u)&=\sum_{n=0}^\infty\frac{n^42^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}u^n\\ &=\frac{6u^3-53u^2+115u-81}{2u(u-2)^3}\\ &\qquad+\frac{u^4-23u^3+89u^2-142u+81}{u^2(u-2)^3\sqrt{\frac{2}{u}-1}}\cdot\arcsin\left(\sqrt{\frac{u}{2}}\right)\tag{12}\\ &=\frac{1}{30}u+\frac{16}{105}u^2+\frac{9}{35}u^3+\frac{1024}{3465}u^4+\frac{2500}{9009}u^5+\cdots\\ \end{align*}

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Building blocks at $u=1$

We have now all the building blocks we need and derive from (8) to (12) some nice identities by setting $u=1$ and noting that $\arcsin\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$.

\begin{align*} A(1)&=\sum_{n=0}^\infty\frac{2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=1-\frac{1}{4}\pi\\ \left.(uD_u)A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n2^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=-\frac{3}{2}+\frac{1}{2}\pi\\ \left.(uD_u)^2A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^22^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=\frac{5}{2}-\frac{3}{4}\pi\\ \left.(uD_u)^3A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^32^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=-\frac{7}{2}+\frac{5}{4}\pi\\ \left.(uD_u)^4A(u)\right|_{u=1}&=\sum_{n=0}^\infty\frac{n^42^{n-1}}{(2n+3)(2n+1)\binom{2n}{n}}=\frac{13}{2}-\frac{3}{2}\pi \end{align*}

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OPs series:

Since \begin{align*} (n^2-n\pi+1)(n^2+n-1)=n^4+(1-\pi)n^3-\pi n^2+(1+\pi)n-1 \end{align*}

We obtain by putting all together in OPs series \begin{align*} \sum_{n=0}^{\infty}&{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}\\ &=\sum_{n=0}^{\infty}{2^n(n^4+(1-\pi)n^3-\pi n^2+(1+\pi)n-1)\over (2n+1)(2n+3){2n\choose n}}\\ &=2\left.(uD_u)^4A(u)\right|_{u=1}+2(1-\pi)\left.(uD_u)^3A(u)\right|_{u=1}\\ &\qquad-2\pi\left.(uD_u)^2A(u)\right|_{u=1}+2(1+\pi)\left.(uD_u)A(u)\right|_{u=1}-2A(1)\\ &=2\left(\frac{13}{2}-\frac{3}{2}\pi\right)+2(1-\pi)\left(-\frac{7}{2}+\frac{5}{4}\pi\right) -2\pi\left(\frac{5}{2}-\frac{3}{4}\pi\right)\\ &\qquad+2(1+\pi)\left(-\frac{3}{2}+\frac{1}{2}\pi\right)-2\left(1-\frac{1}{4}\pi\right)\\ &=1 \end{align*} and the claim follows.

Note: The proof of (1) is Theorem 1 in B. Sury's paper Identities Involving Reciprocals of Binomials. The small trick can be found e.g. in the proof of Theorem 2.1 in Sums of reciprocals of the central binomial coefficients by R. Sprugnoli.

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  • $\begingroup$ Q.E.D. @Markus Scheuer. This is what the OP really looking for step by step. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 13 '16 at 20:52
  • $\begingroup$ @Chinacat: You're welcome! Good to see the answer is useful. :-) $\endgroup$ – Markus Scheuer Jun 13 '16 at 20:55
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    $\begingroup$ it is more than useful, it is a very good knowledge you have shown to me. If I understand this properly, hopefully I can solve my other similar series problems. Thank you again. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 13 '16 at 21:01
  • $\begingroup$ @Chinacat: This was high speed! :-) Many thanks for accepting my answer and granting the bounty. $\endgroup$ – Markus Scheuer Jun 14 '16 at 6:30
  • $\begingroup$ @Chinacat: Some references added at the end of my answer which might be helpful. Regards, $\endgroup$ – Markus Scheuer Jun 14 '16 at 12:19
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Through Euler's beta function you may prove that: $$ \forall x\in(0,4), \qquad f(x)=\sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}} \tag{1}$$ (have also a look at this question), then by computing $\int x^2 f(x^2)\,dx$ we get: $$ \forall x\in(0,2),\quad g(x)=\sum_{n\geq 0}\frac{x^{2n}}{(2n+1)(2n+3)\binom{2n}{n}}=\frac{4 \left(x \sqrt{4-x^2}-(4-x^2)\arcsin\frac{x}{2}\right)}{x^3 \sqrt{4-x^2}}\tag{2}$$ that is more than enough to prove your claim.

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  • $\begingroup$ This is not going into my head! $\endgroup$ – gymbvghjkgkjkhgfkl Jun 7 '16 at 18:33
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$$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}$$

Using Beta function, we can express $${2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}} = {(n^2-n\pi+1)(n^2+n-1)\over {n+1}}\int_{t=0}^1 T^{n+1} dt,$$ where $T =2t(1-t)$. To verify, see that $$\int_{t=0}^1 T^{n+1} dt = 2^{n+1}\beta(n+2,n+2) = 2^{n+1} \frac{(n!)^2 (n+1)^2}{(2n+3)(2n+2)(2n+1)(2n)!} = 2^{n} \frac{ (n+1)}{(2n+3)(2n+1)\binom{2n}{n}}.$$

We further have $$(n^2-n\pi+1)(n^2+n-1) = n^4 +n^3 -n^3\pi-n^2\pi +n(\pi+1)-1.$$ Thus,

$${2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}} = \int_{t=0}^1T^{n+1}\left(n^3 -\pi n^2+(\pi+1)-\frac{\pi+2}{n+1}\right)dt. $$

Thus, $$S= \int_{t=0}^1\sum_{n=0}^\infty T^{n+1}\left(n^3 -\pi n^2+(\pi+1)-\frac{\pi+2}{n+1}\right)dt$$

$$S= \int_{t=0}^1\left(\sum_{n=0}^\infty T^{n+1}n^3 -\pi \sum_{n=0}^\infty T^{n+1}n^2+(\pi+1)\sum_{n=0}^\infty T^{n+1}-\sum_{n=0}^\infty T^{n+1}\frac{\pi+2}{n+1}\right)dt$$

$$S= \int_{t=0}^1 \left(\frac{T^2(1+4T+T^2)}{(1-T)^4} -\pi \displaystyle \frac{T^2(1+T)}{(1-T)^3}+(\pi+1)\frac{T}{1-T}+\ln(1-T)({\pi+2})\right)dt.$$

Substitute back $T=2t(1-t)$ and integrate to get the result. Hope this helps. Using http://www.integral-calculator.com/#expr=%282t%281-t%29%29%5E2%2A%281%2B4%2A%282t%281-t%29%29%2B%282t%281-t%29%29%5E2%29%2F%281-%282t%281-t%29%29%29%5E4%20-pi%2A%282t%281-t%29%29%5E2%2A%281%2B%282t%281-t%29%29%29%2F%281-%282t%281-t%29%29%29%5E3%20%2B%20%28pi%2B1%29%2A%282t%281-t%29%29%2F%281-%282t%281-t%29%29%29%2B%28pi%2B2%29%2Alog%281-%282t%281-t%29%29%29&intvar=t&lbound=0&ubound=1&simplify=1, we have the result as 1 where that link will provide all the detailed steps for this integration. Thus, the required solution $S=1$ has been proved.

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