My Math Forum Uniform convergence of a serie

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 May 19th, 2013, 01:06 AM #1 Newbie   Joined: May 2013 Posts: 7 Thanks: 0 Uniform convergence of a serie My earlier post was about to prove the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if $p<-1$ and divergent if $p\geq -1$. So now I have got this problem to show the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$ is uniform convergent as $x \in ]-\infty;p]$ if $p<-1$. If I am trying ti use the Weierstrass Test I get annoyed when I could not continue what to do the next step when it was wrong. Could you please show me what to do? What I did was, $\left | \frac{\mathrm{ln}(n)^{x}}{n} \right |=\left | \mathrm{ln}(n)^x \right |\left | \frac{1}{n} \right |\leq \left | \mathrm{ln}(n) \right |^{x}\left | \frac{1}{n} \right |\leq \left | \frac{1}{n} \right |=\frac{1}{n}$. Since $\sum \frac{1}{n}$ is not convergent, the abovementioned series couldn't be uniform convergent. Which is why I did something wrong.
May 19th, 2013, 08:23 AM   #2
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Re: Uniform convergence of a serie

Quote:
 Originally Posted by Noworry My earlier post was about to prove the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{p}}{n}$ is convergent if $p<-1$ and divergent if $p\geq -1$. So now I have got this problem to show the series $\sum_{n=2}^{\infty}\frac{\left (\mathrm{ln}\left ( n \right ) \right )^{x}}{n}$ is uniform convergent as $x \in ]-\infty;p]$ if $p<-1$. If I am trying ti use the Weierstrass Test I get annoyed when I could not continue what to do the next step when it was wrong. Could you please show me what to do? What I did was, $\left | \frac{\mathrm{ln}(n)^{x}}{n} \right |=\left | \mathrm{ln}(n)^x \right |\left | \frac{1}{n} \right |\leq \left | \mathrm{ln}(n) \right |^{x}\left | \frac{1}{n} \right |\leq \left | \frac{1}{n} \right |=\frac{1}{n}$. Since $\sum \frac{1}{n}$ is not convergent, the abovementioned series couldn't be uniform convergent. Which is why I did something wrong.
We have already prove that $\sum \frac{\ln^p (n)}{n}$ is convergent for $(-\infty , -1 )$

To prove that $\sum \frac{\ln^x (n)}{n}$ is uniformly convergent let us choose some fixed value $r$ such that $x so we have the following

$\left |\frac{\ln^x (n)}{n} \right | < \frac{\ln^r (n)}{n}$ .

Hence the series is uniformly convergent on $\left(- \infty , x ] \,\, \forall x

May 20th, 2013, 06:46 AM   #3
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Re: Uniform convergence of a serie

Quote:
 Originally Posted by zaidalyafey We have already prove that $\sum \frac{\ln^p (n)}{n}$ is convergent for $(-\infty , -1 )$ To prove that $\sum \frac{\ln^x (n)}{n}$ is uniformly convergent let us choose some fixed value $r$ such that $x so we have the following $\left |\frac{\ln^x (n)}{n} \right | < \frac{\ln^r (n)}{n}$ . Hence the series is uniformly convergent on $\left(- \infty , x ] \,\, \forall x
We know that the earlier series converges as $p \in ]-\infty,-1]$. So we call another value r that contains in the same interval, that is $r\subset p]$. If we look away from the enequality, so this series $\frac{\left (\mathrm{ln}(n) \right )^r}{n}$ looks similar to the first series if we let r = p. Unfortunately the interval of its convergence would exactly be same, so it would be wrong. BUT if we look at this enequality, the convergence of interval as you mentioned $x would be true. Could you please elaborate a bit more because I still feel like I don't understand it completely {I still doubt}?

 May 20th, 2013, 07:33 AM #4 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Uniform convergence of a serie Hopefully an example will clear the doubt Prove that $\sum^{\infty}_{n=2} \frac{\ln^x (n)}{n}$ is uniformly convergent on $x \in ]-\infty , -3]$ since we know that $\sum^{\infty}_{n=1} \frac{\ln^p (n) }{n}$ is convergent forall $p \in ]-\infty,-1[$ , it must be convergent for $p= -2$ , right ? So let us now apply the M-test $\left| \frac{\ln^x (n)}{n} \right |\, < \, \frac{1}{n \, \ln^{2} (n) } \,\,\,\, \forall \,\,\, x \in ]-\infty , -3]$ Hence $\sum^{\infty}_{n=2} \frac{\ln^x (n)}{n}$ is uniformly convergent on $x \in ]-\infty , -3]$ Now , can we generalize that for $x \in ]-\infty , q]\,\, \forall \,\, q < -1$

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