Hello,

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

Thanks!

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- Feb 3rd 2010, 06:01 AMMoshoFunction 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, 09:36 AMgirdav
Take $\displaystyle f_n\left(x\right) =\frac{\left(-1\right)^n}n$ for $\displaystyle x$, for example, in $\displaystyle \left[0,1\right]$. The function is constant so the convergence is uniform and it doesn't converge absolutely.

- Feb 3rd 2010, 09:44 AMMosho
Thanks, but when I say series I mean http://upload.wikimedia.org/math/c/d...3e074ac2bc.png

- Feb 3rd 2010, 09:53 AMDrexel28
That's what he gave you. Let $\displaystyle g(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$. This converges uniformly to $\displaystyle \ln\left(\frac{1}{2}\right)$ for a number of reasons...the easies to just state is Abel's Uniform Convergence Test -- from Wolfram MathWorld with $\displaystyle a_n=\frac{(-1)^n}{n},f_n(x)=1$.

- Feb 3rd 2010, 10:02 AMMosho
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, 10:40 AMMosho
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, 12:21 PMDrexel28
- Feb 3rd 2010, 12:35 PMMosho
Sorry, conditionally convergent.