# Determine the convergence, both pointwise and uniform

• Mar 23rd 2011, 05:22 PM
wopashui
Determine the convergence, both pointwise and uniform
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!
• Mar 23rd 2011, 07:24 PM
TheEmptySet
Quote:

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.
• Mar 23rd 2011, 08:57 PM
Also sprach Zarathustra
Perhaps a hint...:

• Mar 24th 2011, 07:31 PM
wopashui
Quote:

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
• Mar 25th 2011, 08:11 AM
TheEmptySet
Quote:

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

$\displaystyle \bigg|\frac{1}{e^x} \bigg| < 1$
• Mar 25th 2011, 08:33 AM
Plato
Quote:

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: $S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x}$
• Mar 25th 2011, 07:30 PM
wopashui
Quote:

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?
• Mar 25th 2011, 07:32 PM
wopashui
Quote:

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?
• Mar 25th 2011, 08:01 PM
TheEmptySet
Quote:

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?
• Mar 25th 2011, 09:21 PM
wopashui
Quote:

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 $e^0=1$?
• Mar 25th 2011, 10:39 PM
TheEmptySet
Quote:

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.