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Math Help - Determine the convergence, both pointwise and uniform

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    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!
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    Quote Originally Posted by wopashui View Post
    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.
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    Perhaps a hint...:

    Your series is telescopic.
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    Quote Originally Posted by TheEmptySet View Post
    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
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    Quote Originally Posted by wopashui View Post
    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
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    Quote Originally Posted by wopashui View Post
    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}
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    Quote Originally Posted by Plato View Post
    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?
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    Quote Originally Posted by TheEmptySet View Post
    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?
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    Quote Originally Posted by wopashui View Post
    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?
    Last edited by TheEmptySet; March 25th 2011 at 10:38 PM. Reason: typo
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    Quote Originally Posted by TheEmptySet View Post
    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?
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    Quote Originally Posted by wopashui View Post
    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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