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In the problem the sum is given as $$\sum\limits_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}$$ and then when I try to solve it using Gauss's test I get $$\frac{a_{n}}{a_{n+1}}=\frac{2n+2}{2n+1}$$ but in the solution there is given: $$\frac{a_{n}}{a_{n+1}}=\frac{2n}{2n-1}$$

my reasoning was:

$$\frac{a_{n}}{a_{n+1}}=\frac{\frac{(2n-1)!!}{(2n)!!}}{\frac{(2n+1)!!}{(2n+2)!!}}=\frac{1\cdot3\cdot...\cdot(2n-1) \times 2\cdot4\cdot...\cdot2n\cdot(2n+2)}{2\cdot4\cdot...\cdot2n \times 1\cdot3\cdot(2n-1)\cdot(2n+1)}=\frac{2n+2}{2n+1}$$ I believe that I made a mistake, but I don't know where?

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  • $\begingroup$ What is it you are trying to do? $\endgroup$
    – Ron Gordon
    Nov 3, 2015 at 17:55
  • $\begingroup$ Your calculation of the ratio $\frac{a_n}{a_{n+1}}$ is correct. The one you quote seems to have an index off by one issue. The ratio is unfortunately not directly useful in dealing with the issue of convergence. $\endgroup$
    – André Nicolas
    Nov 3, 2015 at 17:58
  • $\begingroup$ To see is my reasoning true or not, because from that point I can solve the rest of the problem... $\endgroup$
    – mlata
    Nov 3, 2015 at 17:58
  • $\begingroup$ As an aside, $\displaystyle\sum_{n=0}^\infty \bigg[\frac{(2n-3)!!}{(2n)!!}\bigg]^2~=~\frac4\pi$ $\endgroup$
    – Lucian
    Nov 3, 2015 at 20:41

1 Answer 1

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$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \frac{1}{4^n}\binom{2n}{n} $$ but the RHS behaves like $\frac{1}{\sqrt{\pi n}}$, hence the series is diverging. Also without the exact asymptotics:

$$ \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=1}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) \geq \frac{1}{4n}.$$

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