# Thread: Determine the convergence, both pointwise and uniform

1. ## Determine 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!

2. Originally Posted by wopashui
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!
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.

3. Perhaps a hint...:

4. Originally Posted by TheEmptySet
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.

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

5. 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 $\displaystyle |r| < 1$ so in this cases whe

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

6. 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 $\displaystyle \displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} }$.
Now you have a rather simple sequence: $\displaystyle S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x}$

7. 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 $\displaystyle \displaystyle S_N = \sum\limits_{n =1 }^N {\left[ {e^{ - nx} - e^{ - \left( {n + 1} \right)x} } \right] = e^{ - x} - e^{ - x\left( {N + 1} \right)} }$.
Now you have a rather simple sequence: $\displaystyle S_N = \left( {e^{ - x} - e^{ - x\left( {N + 1} \right)} } \right) \to e^{ - x}$
sorry, i dun understand what you are doing here, why $\displaystyle e^{-nx}$ becomes $\displaystyle e^{-x}$, what did n go?

8. Originally Posted by TheEmptySet
Use the Weierstrass M-test. If will converge uniformly and absolutely when $\displaystyle |r| < 1$ so in this cases whe

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

9. Originally Posted by wopashui
so if we $\displaystyle 1/e^x$ will bounded by 1, so if we take $\displaystyle M_n = 1^n$, which diverges, then the M test won't work, did I make mistake here?
That is exactly the point!

If you let $\displaystyle x=0$ 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?

10. Originally Posted by TheEmptySet
That is exactly the point!

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

$\displaystyle \displaystyle \sum_{n=1}^{\infty}[e^{-nx}-e^{-(n+1)x}]=\sum_{n=1}^{\infty}[0-0]=0$ 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?
so the series is pointwise convergent but not uniformly convergent, right? btw, isn't $\displaystyle e^0=1$?

11. Originally Posted by wopashui
so the series is pointwise convergent but not uniformly convergent, right? btw, isn't $\displaystyle e^0=1$?
yes, there was a typo above and I fixed it.