# Function series which converges uniformly but not absolutely

• Feb 3rd 2010, 07:01 AM
Mosho
Function series which converges uniformly but not absolutely
Hello,

I'm new here and would like some help finding a function series qualifying the conditions in the title.

Thanks!
• Feb 3rd 2010, 10:36 AM
girdav
Take $f_n\left(x\right) =\frac{\left(-1\right)^n}n$ for $x$, for example, in $\left[0,1\right]$. The function is constant so the convergence is uniform and it doesn't converge absolutely.
• Feb 3rd 2010, 10:44 AM
Mosho
Thanks, but when I say series I mean http://upload.wikimedia.org/math/c/d...3e074ac2bc.png
• Feb 3rd 2010, 10:53 AM
Drexel28
Quote:

Originally Posted by Mosho
Thanks, but when I say series I mean http://upload.wikimedia.org/math/c/d...3e074ac2bc.png

That's what he gave you. Let $g(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$. This converges uniformly to $\ln\left(\frac{1}{2}\right)$ for a number of reasons...the easies to just state is Abel's Uniform Convergence Test -- from Wolfram MathWorld with $a_n=\frac{(-1)^n}{n},f_n(x)=1$.
• Feb 3rd 2010, 11:02 AM
Mosho
My bad then. I had an exam today and what you guys said is what I wrote, but most people said I was wrong.

Thanks alot :)
• Feb 3rd 2010, 11:40 AM
Mosho
I have another question:

Is this statement true/false?

"The sum of a absolutely convergent series and a semi convergent series is a semi convergent series"

Thanks again!
• Feb 3rd 2010, 01:21 PM
Drexel28
Quote:

Originally Posted by Mosho
I have another question:

Is this statement true/false?

"The sum of a absolutely convergent series and a semi convergent series is a semi convergent series"

Thanks again!

What does semi-convergent mean?
• Feb 3rd 2010, 01:35 PM
Mosho
Sorry, conditionally convergent.