Sum of series with changing ratio I have a series of the type
$$\frac{1}{2} + \frac{1}{2} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} + \dots $$
The series has a changing ratio $= \frac{2k-1}{2k}$.
The ratio in itself is increasing but always less than $1$.
Q. Is the sum of the series finite? What is the sum of the series?
I have programmed it in Matlab, but the result has not converged even after $10^{13}$ terms.
The difference keeps on decreasing, but the rate of decrease in difference is also decreasing.
Any reference to literature is also appreciated.
Thank you for your time !
 A: We know that $(1+x)^n= 1+nx+n(n-1)/2 .x^2+n(n-1)(n-2)/6 .x^3 .... $
for any rational index $n$. Hence, putting$n=-1/2$ gives
$(1+x)^{-1/2}= 1-x/2+(1/2).(3/4)x^2-(1/2).(3/4).(5/6)x^3+...$
Now, we take $\lim_{x\to-1^+}$ both sides and analyse them one by one
When $x$ tends to $-1$ from right side, we see that the right hand side becomes
$1+1/2+(1/2).(3/4)+(1/2).(3/4).(5/6)....$ which is one more than what is required.
And the left hand side tends to positive infinity as $x$ approaches $-1$ from right side. Hence the required sequence diverges and the sum tends to positive infinity.
A: By Stirling's approximation, $n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n$. On the other hand,\begin{align}\frac12\times\frac34\times\cdots\times\frac{2n-1}{2n}&=\frac{(2n-1)!!}{(2n)!!}\\&=\frac{(2n-1)!!\times2\times4\times\cdots\times(2n)}{2\times4\times\cdots\times(2n)\times2\times4\times\cdots\times(2n)}\\&=\frac{(2n)!}{2^{2n}(n!)^2}.\end{align}Putting all this together, you get that\begin{align}\frac12\times\frac34\times\cdots\times\frac{2n-1}{2n}&\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2^{2n}2\pi n\left(\frac ne\right)^{2n}}\\&=\frac1{\sqrt{\pi n}}.\end{align}Since the series $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{\pi n}}$ diverges, then so does your series.
A: You have
$$a_n=\frac{\prod_{k=1}^n (2k-1) }{\prod_{k=1}^n (2k) }=\frac{\Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }\, \Gamma (n+1)}$$
$$S_p=\sum_{n=1}^p a_n=\frac{2 \Gamma \left(p+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (p+1)}-1$$
Using Stirling approximation
$$S_p=2 \sqrt{\frac p \pi}\Bigg[1+\frac{3}{8 p}+O\left(\frac{1}{p^2}\right) \Bigg]-1$$ Trying for $p=10^4$ this gives, as an approximation,
$111.842148131$ while the exact value is
$111.842148067$
A: Here is a quite elementary way to handle this series:
Set
$$a_n :=\frac 12\cdot \frac 34 \cdots \frac{2n-1}{2n} = \prod_{k=1}^n \frac{2k-1}{2k}$$
Now, set $c_1 = \frac 12$ and for $n \geq 2$
$$c_n := \frac 12\cdot \frac 23 \cdots \frac{2(n-1)}{2n-1}=\frac 12\cdot \prod_{k=2}^{n} \frac{2(k-1)}{2k-1}$$
You can quickly check that
$$c_n < a_n \text{ for } n>1$$
Hence, for $n>1$ you have
$$c_na_n = \frac 14 \prod_{k=2}^{n} \frac{2(k-1)}{2k} \stackrel{telescope}{=}  \frac 14 \frac{2}{2n} = \frac 1{4n} < a_n^2$$
So, for $n>1$ you have
$$a_n > \frac 1{2\sqrt n}$$
Since $\sum_{n=1}^{\infty}\frac 1{2\sqrt n} = \infty$, it follows that $\sum_{n=1}^{\infty}a_n = \infty$, too.
