How to sum this series for π2 directly? π2=∑k=0∞k!(2k+1)!

Question

How to sum this series for π2 directly?  π2=∑k=0∞k!(2k+1)!

Frauental91 · Accepted Answer

First,  (2k+1)≠(2k+1)(2k−1)  =(2k+1)!(2k)(2(k−1))…2(1)=(2k+1)!2kk!  So your sum can be rewritten as  ∑k=0∞k!k!2k(2k+1)!=∑k=0∞2k(2k+1)(2kk)  Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For example, this paper by Sprugnoli gives the ordinary generating function of ak=4k(2k+1)(2kk)−1 to be  A(t)=1tt1−tarctan⁡t1−t  Subbing in t=12 says that  ∑k=0∞2k(2k+1)(2kk)=2arctan⁡(1)=π2

primenamaqm · Accepted Answer

We can prove this identity, as well as the corresponding power series identities by using a relation with the Beta function.  ∑k=0∞k!k!2k(2k+1)!  Using induction or a Beta Function identity, we can show that  ∫01xk(1−x)k=k!k!(2k+1)!  Hence your sum becomes  ∑k=0∞2k∫01xk(1−x)k=∫01(∑k=0∞2kxk(1−x)k)dx  Notice that since 0≤x≤1, x(1−x)≤14 and the series converges absolutely. Summing gives  =∫0111−2x(1−x)dx=∫011x2+(1−x)2dx  Substituting u=1x, and then v=u−1, we see that this integral is equal to  ∫1∞11+(u−1)2du=∫0∞11+v2dv=π2  as desired

RizerMix · Accepted Answer

∫0π2sin2k+1⁡(x)dx=2k2k+12k−22k−1...23 (1) =12k+14k((2k),(k)) Using (1), mu sum becomes ∑k=0∞k!(2k+1)!=∑k=0∞2k(2k+1)((2k),(k)) =∑k=0∞∫0π22(sin⁡(x)2)2k+1dx =2∫0π2(sin⁡(x)2)1−(sin⁡(x)2)2dx =∫0π22sin⁡(x)2−sin2⁡(x)dx =∫π202dcos⁡(x)1+cos2⁡(x) =π2

How to sum this series for \pi/2 directly? \frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!}

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