2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What is it you are trying to do? $\endgroup$ – Ron Gordon Nov 3 '15 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 '15 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 '15 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 '15 at 20:41
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.