Proof the convergence of a gamma sum

In summary: The series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges... all right?... then the series $\displaystyle \sum_{n=1}^{\infty} \frac{a_{n}}{n}$ if for all n is $a_{n}>1$ also converges... all right?...In summary, the series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges if and only if $a_{n}>1$.
  • #1
alyafey22
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How to prove the convergence or divergence of ?

$$\sum^{\infty}_{n=1}\frac{\Gamma(n+\frac{1}{2})}{n\Gamma{(n+\frac{1}{4})}}$$
 
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  • #2
ZaidAlyafey said:
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

$\chi$ $\sigma$
 
  • #3
chisigma said:
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 ?
 
  • #4
ZaidAlyafey said:
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 $\chi$ $\sigma$
 
  • #5
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 .
 
  • #6
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

$\chi$ $\sigma$
 

Related to Proof the convergence of a gamma sum

1. What is the gamma sum?

The gamma sum, or gamma function, is a mathematical function that extends the concept of factorial to real and complex numbers. It is denoted by the symbol Γ and is defined as Γ(z) = ∫0∞ t^(z-1) * e^(-t) dt.

2. What does it mean for a gamma sum to converge?

Convergence of a gamma sum means that as the number of terms in the sum increases, the sum approaches a finite value. In other words, the sum does not diverge to infinity.

3. How can we prove the convergence of a gamma sum?

One method to prove the convergence of a gamma sum is by using the ratio test, which compares the ratio of consecutive terms in the sum to a limit. If the limit is less than 1, then the sum converges. Another method is to use the integral test, which compares the sum to an integral and determines convergence based on the behavior of the integral.

4. What factors affect the convergence of a gamma sum?

The convergence of a gamma sum is affected by the value of the parameter z in the gamma function. If z is a negative integer, the sum will diverge. Additionally, the rate of convergence may also be affected by the value of z and the number of terms in the sum.

5. Can the convergence of a gamma sum be proven for all values of z?

No, the convergence of a gamma sum can only be proven for certain values of z. For example, the sum will always converge for positive real numbers, but may not converge for negative real numbers or complex numbers with non-zero imaginary parts. The convergence can also depend on the specific method used to prove it.

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