4

Convergence of $$\sum_{n=2}^\infty \frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$$

Well, I have tried almost everything. D'Alembert's criterion doesn't work because the limit is 1. I have tried to bound $\frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}$ in several ways but every bound diverges. Any hint?

P.S: I know this series converges to $\frac{1}{2}$

Did
  • 279,727
davidaap
  • 708
  • Stirling's formula? – Angina Seng Oct 16 '17 at 05:39
  • 1
    You might want to use $(2n-2)!=[1\cdot3\cdots(2n-3)][2\cdot4\cdots(2n-2)=[1\cdot3\cdots(2n-3)]2^{n-1}(n-1)!$. – Gerry Myerson Oct 16 '17 at 05:41
  • I already used that and didn't help. – davidaap Oct 16 '17 at 05:51
  • 1
    For a low-tech approach, you might check that, for every $n\geqslant2$, $$\frac{2n-3}{2n}\leqslant\left(\frac{n}{n+1}\right)^{3/2}$$ by verifying that $$(2n-3)^2(n+1)^3\leqslant n^3(2n)^2$$ and deduce that $$\frac{1\cdot 3\cdot \cdot \cdot (2n-3)}{2^n n!}=\frac12\prod_{k=2}^n\frac{2k-3}{2k}\leqslant\frac12\prod_{k=2}^n\left(\frac{k}{k+1}\right)^{3/2}=\frac{\sqrt2}{(n+1)^{3/2}}$$ from which the convergence of the series follows. – Did Oct 16 '17 at 08:31

2 Answers2

2

You can use Gauss test (or Rabee's test, the latter is weaker but still suffices), I will use the Gauss test:

If $u_n >0$ and satisfies $$\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(\frac{1}{n^r})$$ for some $r>1$, then $\sum u_n$ converges iff $h>1$.

In your case, $$u_n = \frac{1\cdot 3\cdots (2n-3)}{2^n n!} = \frac{(2n-2)!}{2^{2n-1}n!(n-1)!}$$ we easily obtain $$\frac{u_n}{u_{n+1}} = \frac{2n+2}{2n-1} = 1+\frac{3}{2n-1} = 1+\frac{3}{2n}+O(\frac{1}{n^2})$$ so $\sum u_n$ converges.

Alternatively, you can use the Stirling formula $$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$ but I think this is an overkill.

The value of the sum can be derived from the following series (after integrating and shifting terms): $$\sum_{n=0}^\infty \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}$$

pisco
  • 18,983
0

$$\sum_{n\geq 2}\frac{(2n-3)!!}{2^n n!} = \sum_{n\geq 1}\frac{(2n-1)!!}{2^{n+1}(n+1)!}=\sum_{n\geq 1}\frac{(2n)!}{2^{2n+1}n!(n+1)!}=\sum_{n\geq 1}\binom{2n}{n}\frac{1}{4^n(2n+2)} $$ and since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$, the given series is convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{1}{n\sqrt{n}}$, which is convergent by the $p$-test. The given series equals $\frac{1}{2}$ by creative telescoping or probabilistic arguments: have a look at this similar question.

Jack D'Aurizio
  • 353,855