Why does $1+(\frac12)(\frac13)+(\frac{1\cdot3}{2\cdot4})(\frac15)+(\frac{1\cdot3\cdot5}{2\cdot4\cdot6})(\frac17)+\cdots$ converge?

Question

I am a newbie to series. From the 3rd page of the MIT OpenCourseWare Single Variable Calculus "Final Review" lecture notes (PDF link via mit.edu), we know that

$$1 + \left(\frac{1}{2} \right) \left(\frac{1}{3} \right) + \left( \frac{1\cdot3}{2\cdot4} \right) \left(\frac{1}{5}\right) + \left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)\left(\frac{1}{7}\right)+\cdots$$

converges to $\dfrac{\pi}{2}$. The result is achieved by using the Taylor series of $(1+u)^{-\frac{1}{2}}$.

The lecture note left a problem:

How to prove the series' convergence by using L'Hospital rule?

I am not sure how L'Hospital rule helps here. I tried to prove that the $\frac{1}{n^2}$ is the upper limit of each term $\left(\frac{1*3*5*...*(2n-1)}{2*4*6*...*(2n)}\right)\left(\frac{1}{2n+1}\right)$, but got no luck.

Any idea?

Answer 1

I can see a few issues in your post and I will briefly address them first before tackling convergence.

The series is obtained by using term by term integration on the Taylor series of $(1-u^2)^{-1/2}$ on interval $[0,1]$ and its value is $\pi/2$.

The more serious issue is that we don't prove the convergence / divergence of a series using L'Hospital's Rule. L'Hospital's Rule is a tool which can be applied under certain circumstances to evaluate limit of functions of a real variable.

The convergence/divergence of a series is determined in general using a set of well known criteria usually called tests for convergence.

In the case of series in question we can note that if $a_n$ denotes the $n$-th term of the series (ignoring first term) then $$\frac{a_{n}}{a_{n+1}}=\frac{2(n+1)(2n+3)}{(2n+1)^2}=1+\frac{6n+5}{(2n+1)^2}$$ so that $$n\left(\frac{a_n} {a_{n+1}}-1\right)\to\frac{3}{2}$$ The above limit is greater than $1$ and hence the series converges by Raabe's test.

Answer 2

Hint: $$\frac{1}{2\sqrt{n}}\leqslant \frac{1}{2}\cdot\frac{3}{4}\cdot \frac{5}{6}\cdots \frac{2n-1}{2n}\lt \frac{1}{\sqrt{2n}}$$