I have to study the convergence of the following series:

$$\frac{1}{4} + \frac{1\cdot 9}{4\cdot 16} + \frac{1\cdot 9 \cdot 25}{4\cdot 16\cdot 36}+\ldots$$

I ended up with $$\frac{a_{n+1}}{a_n} = \frac{(2n+1)^2}{(2n+2)^2}$$

But couldn't move from here. Since by ratio test it's limit is 1 and we can't conclude anything.
Any tips?


You may notice that: $$ a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)^{2n}\,dx \tag{1}$$ from which: $$ f(x)=\sum_{n\geq 1}\frac{(2n-1)!!}{(2n)!!}\,x^{2n} = -1+\frac{1}{\sqrt{1-x^2}}.\tag{2} $$ Assuming that your series is convergent, we have $$ S = \sum_{n\geq 1}\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \frac{2}{\pi}\int_{0}^{\pi/2}f(\cos x)\,dx =-\frac{2}{\pi}+\int_{0}^{\pi/2}\frac{dx}{\sin x}\tag{3}$$ but the last integral is divergent, and so it is $S$. An alternative approach comes from $$ \frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}\tag{4} $$ granting that the original series is divergent by comparison with the harmonic series.

An interesting identity related with elliptic integrals is the following: $$ \sum_{n\geq 1}\left(\frac{(2n-1)!!}{(2n)!!}\right)^{\color{red}{3}} = \color{red}{-1+\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}}.$$


You can use Raabe's Test which is:

n[ {a(n)/a(n+1)} -1 ] =L as n ->infinity.

If R>1 its convergent ( because it's the inverse of the ratio in the ratio test ). If R<1 then divergent. If R is equal to 1 then you will have to use De Morgan Bertrand Test.

  • $\begingroup$ I couldn't apply the Raabe's Test - I mean I couldn't get this into that form $\endgroup$ – user346936 Feb 23 '17 at 12:22
  • $\begingroup$ @Xenidia You have got [ a(n+1)/a(n) ]. Just reverse it. It will be [ { a(n)/a(n+1) }] . Subtract 1 and multiply by n and find the limit as n tends to infinity. That's it ! $\endgroup$ – Shashaank Feb 23 '17 at 12:25

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