# Proof the convergence of a gamma sum

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
How to prove the convergence or divergence of ?

$$\sum^{\infty}_{n=1}\frac{\Gamma(n+\frac{1}{2})}{n\Gamma{(n+\frac{1}{4})}}$$

#### chisigma

##### Well-known member
How to prove the convergence or divergence of ?

$$\sum^{\infty}_{n=1}\frac{\Gamma(n+\frac{1}{2})}{n\Gamma{(n+\frac{1}{4})}}$$
For x>2 $\Gamma(x)$ is increasing with x, so that is $\displaystyle \frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+\frac{1}{4})} > 1$ so that...

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
For x>2 $\Gamma(x)$ is increasing with x, so that is $\displaystyle \frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+\frac{1}{4})} > 1$ so that...

Kind regards

$\chi$ $\sigma$
I don't get what you are implying ?

#### chisigma

##### Well-known member
I don't get what you are implying ?
The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges... all right?... then the series $\displaystyle \sum_{n=1}^{\infty} \frac{a_{n}}{n}$ if for all n is $a_{n}>1$ also diverges... all right?...

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
all right that is clear , actually I made a mistake I intended something different :

$$\sum^{\infty}_{n=1} \frac{\Gamma{(n+\frac{1}{4})}}{n\Gamma{(n+\frac{1}{2})}}$$

Sorry for the typo .

#### chisigma

##### Well-known member
All right!... first we write the series as $\displaystyle \sum_{n=1}^{\infty} a_{n}$ and the we remember one of the basic properties of the Gamma function…

$\displaystyle \Gamma (x+n)= \Gamma (x)\ x\ (x+1)\ ...\ (x+n-1)$ (1)

... that permits us, setting...

$\displaystyle b_{n}= \frac{\Gamma(n+\frac{1}{4})}{\Gamma(n+\frac{1}{2})}$ (2)

... the recursive relation...

$\displaystyle b_{n+1}= b_{n}\ \frac{n + \frac{1}{4}}{n + \frac{1}{2}}$ (3)

Because is $\displaystyle a_{n}=\frac{b_{n}}{n}$ from (3) we derive...

$\displaystyle a_{n+1}= a_{n}\ \frac{n + \frac{1}{4}}{n + \frac{1}{2}}\ \frac{n}{n+1} \implies \frac{a_{n}}{a_{n+1}} = \frac{n^{2} + \frac{3}{2} n + \frac{1}{2}}{n^{2}+ \frac{1}{4} n}$ (4)

Now we can use the so called 'Raabe test' that extablishes that if for n 'large enough' the following relation...

$\displaystyle c_{n}= n\ (\frac {a_{n}}{a_{n+1}} -1) \ge h >1$ (5)

... is verified then the series converges. From (4) and (5) it is easy to verify that $\displaystyle \lim_{n \rightarrow \infty} c_{n}= \frac{5}{4}$ so that the convergence is proved...

Kind regards

$\chi$ $\sigma$