Determine the convergence, both pointwise and uniform, on [0,1] for the series

$\displaystyle \sum_{n = 1}^{\infty} [e^{-nx}-e^{-(n+1)x}] $

hint: look at the partial sums!

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- Mar 23rd 2011, 04:22 PMwopashuiDetermine the convergence, both pointwise and uniform
Determine the convergence, both pointwise and uniform, on [0,1] for the series

$\displaystyle \sum_{n = 1}^{\infty} [e^{-nx}-e^{-(n+1)x}] $

hint: look at the partial sums! - Mar 23rd 2011, 06:24 PMTheEmptySet
Note that

$\displaystyle \displaystyle e^{-nx}-e^{-(n+1)x}=\left( \frac{1}{e^{x}}\right)^n\left( 1-e^{-x}\right)$

$\displaystyle \displaystyle \left( 1-e^{-x}\right)\sum_{n=1}^{\infty}\left( \frac{1}{e^{x}}\right)^n$

The series is geometric with $\displaystyle \displaystyle r= \frac{1}{e^{x}}$

So when does the geometric series converge uniformly. - Mar 23rd 2011, 07:57 PMAlso sprach Zarathustra
Perhaps a hint...:

Your series is telescopic. - Mar 24th 2011, 06:31 PMwopashui
- Mar 25th 2011, 07:11 AMTheEmptySet
- Mar 25th 2011, 07:33 AMPlato
No need for geometric series

Did you even bother with the hint given in the OP?

Use partial collapsing sums.

Let $\displaystyle \displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} } $.

Now you have a rather simple sequence: $\displaystyle S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x} $ - Mar 25th 2011, 06:30 PMwopashui
- Mar 25th 2011, 06:32 PMwopashui
- Mar 25th 2011, 07:01 PMTheEmptySet
That is exactly the point!

If you let $\displaystyle x=0$ you get the series

$\displaystyle \displaystyle \sum_{n=1}^{\infty}[e^{-nx}-e^{-(n+1)x}]=\sum_{n=1}^{\infty}[1-1]=0$ so the series converges pointwise to zero there

Now by the geometric series when $\displaystyle x \ne 0$ the series converges to

$\displaystyle \displaystyle 1-e^{-x}\left( \frac{e^{-x}}{1-e^{-x}}\right)=e^{-x}$

so this sum converges to the function

$\displaystyle f(x)=\begin{cases}0 \, , \text{ if } x=0 \\ e^{-x}, \text{ if } x \in (0,1]\end{cases}$

Can a series of uniformly continuous functions converge to a discontinuous function? - Mar 25th 2011, 08:21 PMwopashui
- Mar 25th 2011, 09:39 PMTheEmptySet