Determine the convergence, both pointwise and uniform

**wopashui** · March 23rd 2011, 04:22 PM

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

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

hint: look at the partial sums!

**TheEmptySet** · March 23rd 2011, 06:24 PM

Originally Posted by wopashui

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

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

hint: look at the partial sums!

Note that

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

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

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

So when does the geometric series converge uniformly.

**Also sprach Zarathustra** · March 23rd 2011, 07:57 PM

Perhaps a hint...:

Your series is telescopic.

**wopashui** · March 24th 2011, 06:31 PM

Originally Posted by TheEmptySet

Note that

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

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

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

So when does the geometric series converge uniformly.

we actually have not talked about when does the geometric series converge uniformly, any hints? And what about pointwise convergence

**TheEmptySet** · March 25th 2011, 07:11 AM

Originally Posted by wopashui

we actually have not talked about when does the geometric series converge uniformly, any hints? And what about pointwise convergence

Use the Weierstrass M-test. If will converge uniformly and absolutely when $|r| < 1$ so in this cases whe

$\displaystyle \bigg|\frac{1}{e^x} \bigg| < 1$

**Plato** · March 25th 2011, 07:33 AM

Originally Posted by wopashui

we actually have not talked about when does the geometric series converge uniformly, any hints? And what about pointwise convergence

No need for geometric series
Did you even bother with the hint given in the OP?
Use partial collapsing sums.
Let $\displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} }$ .
Now you have a rather simple sequence: $S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x}$

**wopashui** · March 25th 2011, 06:30 PM

Originally Posted by Plato

No need for geometric series
Did you even bother with the hint given in the OP?
Use partial collapsing sums.
Let $\displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} }$ .
Now you have a rather simple sequence: $S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x}$

sorry, i dun understand what you are doing here, why $e^{-nx}$ becomes $e^{-x}$ , what did n go?

**wopashui** · March 25th 2011, 06:32 PM

Originally Posted by TheEmptySet

Use the Weierstrass M-test. If will converge uniformly and absolutely when $|r| < 1$ so in this cases whe

$\displaystyle \bigg|\frac{1}{e^x} \bigg| < 1$

so if we $1/e^x$ will bounded by 1, so if we take $M_n = 1^n$ , which diverges, then the M test won't work, did I make mistake here?

**TheEmptySet** · March 25th 2011, 07:01 PM

Originally Posted by wopashui

so if we $1/e^x$ will bounded by 1, so if we take $M_n = 1^n$ , which diverges, then the M test won't work, did I make mistake here?

That is exactly the point!

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

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

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

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

so this sum converges to the function

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

Can a series of uniformly continuous functions converge to a discontinuous function?

**wopashui** · March 25th 2011, 08:21 PM

Originally Posted by TheEmptySet

That is exactly the point!

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

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

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

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

so this sum converges to the function

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

Can a series of uniformly continuous functions converge to a discontinuous function?

so the series is pointwise convergent but not uniformly convergent, right? btw, isn't $e^0=1$ ?

**TheEmptySet** · March 25th 2011, 09:39 PM

Originally Posted by wopashui

so the series is pointwise convergent but not uniformly convergent, right? btw, isn't $e^0=1$ ?

yes, there was a typo above and I fixed it.

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