Analyzing Convergent Series w/Supremum & Integral Norms

  • Context: Graduate 
  • Thread starter Thread starter BSCowboy
  • Start date Start date
  • Tags Tags
    Convergent Series
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
BSCowboy
Messages
31
Reaction score
0
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].
when
[tex]f_n(t)=\frac{t^n}{n}[/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:
on Phys.org
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].
 
[tex] \sum_{n=1}^\infty {\frac 1 {n(n+1)} \not \to 0[/tex]

but

[tex] \lim_{n \to \infty} \frac 1 {n(n+1)} \to 0[/tex]
 
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]