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Convergent Series

  1. Mar 3, 2010 #1
    I have been thinking about this problem:
    Determine whether the following series are convergent in [tex]\left(C[0,1],||\cdot ||_{\infty}\right)[/tex] and [tex]\left(C[0,1],||\cdot ||_{1}\right)[/tex].

    In the supremum norm, this seems pretty straightforward, but in the integral norm I am confused since,
    [tex]\left\|\sum\frac{t^n}{n}\right\|_1\leq\sum\left\|\frac{t^n}{n}\right\|_1=\sum\int_0^1\frac{t^n}{n}dt=\sum\left[\frac{t^{n+1}}{n^2+n}\right]_0^1=\sum\frac{1}{n^2+n}<\sum\frac{1}{n^2} [/tex]
    and, I think this converges as [tex]n\rightarrow\infty[/tex], but our instructor said this did not converge, or maybe I heard him incorrectly. So, does this converge? He asked us to show the series is Cauchy and that the limit is not in the space as well. What am I missing?
    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2


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    Could you be more precise as to what is supposed to be converging to what? Partial sums of fn(t) or fn(t) itself as a sequence.
  4. Mar 4, 2010 #3
    Actually, I think I might have figured it out:
    [tex]f_n(t)=\frac{t^n}{n} \text{ in } \left(C[0,1],\|\cdot\|_1\right)[/tex]

    [tex]\|f_n\|_1=\int_0^1f_n(t)dt=\frac{1}{n(n+1)} \text{ and } \sum_{n=1}^{\infty}\frac{1}{n(n+1)}\rightarrow 0[/tex]

    But, because the space [tex]\left(C[0,1],\|\cdot\|_1\right)[/tex] is not complete all we know is that [tex]f_n(t)[/tex] is Cauchy. I was missing the part about the space being complete.
    Also, you can see that limit is not an element of C[0,1].
  5. Mar 5, 2010 #4


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    \sum_{n=1}^\infty {\frac 1 {n(n+1)} \not \to 0


    \lim_{n \to \infty} \frac 1 {n(n+1)} \to 0
  6. Mar 6, 2010 #5
    Also [tex]\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left[ \frac{1}{n} - \frac{1}{n+1} \right] = 1[/tex]
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