# Simple Convergence limit

• December 1st 2011, 11:13 AM
Simple Convergence limit
Hey, I'm trying to show that the following sequence of functions
$f_n(x)=\sum_{k=1}^{n}\frac{1}{(x+k)^2}$
converges uniformly on $\mathbb{R}^+$.

To do that I'm trying to find the limit of the simple convergence of the sequence, I can easily show that the limit for the simple convergence exists by noticing that for every $x\in\mathbb{R}^+$,
$0\leq\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\leq\sum_ {n=1}^{\infty}\frac{1}{n^2}$
and use the convergence of the Riemann Zeta Function. However I can't manage to what the function of the limit is.
• December 1st 2011, 04:07 PM
chisigma
Re: Simple Convergence limit
Quote:

Hey, I'm trying to show that the following sequence of functions
$f_n(x)=\sum_{k=1}^{n}\frac{1}{(x+k)^2}$
converges uniformly on $\mathbb{R}^+$.

To do that I'm trying to find the limit of the simple convergence of the sequence, I can easily show that the limit for the simple convergence exists by noticing that for every $x\in\mathbb{R}^+$,
$0\leq\sum_{n=1}^{\infty}\frac{1}{(x+n)^2}\leq\sum_ {n=1}^{\infty}\frac{1}{n^2}$
and use the convergence of the Riemann Zeta Function. However I can't manage to what the function of the limit is.

In...

http://www.mathhelpforum.com/math-he...-i-188482.html

... it has been demonstrated that...

$\sum_{n=1}^{\infty} \frac{1}{(n+a)\ (n+b)}= \frac{\phi(b)-\phi(a)}{b-a}$ (1)

... where...

$\phi(x)= \frac{d}{dx} \ln x!$ (2)

... being x! the so called 'factorial function'...

$x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt$ (3)

Now from (1) using l'Hopital rule You obtain...

$\sum_{n=1}^{\infty} \frac{1}{(x+n)^{2}} = \lim_{\xi \rightarrow x} \sum_{n=1}^{\infty} \frac{1}{(\xi+n)\ (x+n)} =\lim_{\xi \rightarrow x} \frac{\phi(\xi)-\phi(x)}{\xi-x} = \phi^{'} (x)$ (4)

Kind regards

$\chi$ $\sigma$