Analyzing Convergent Series w/Supremum & Integral Norms

In summary, the conversation discusses the convergence of a series in two different norms, the supremum norm and the integral norm. While it is straightforward to show convergence in the supremum norm, there is confusion about the convergence in the integral norm. The instructor has asked for the series to be shown as Cauchy and for the limit to not be in the given space, but the conversation concludes that the series does converge in the integral norm. However, the space itself is not complete, so we can only say that the series is Cauchy. The conversation also mentions that the limit is not an element of the space and provides an example of this.
  • #1
31
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?
 
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  • #2
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.
 
  • #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].
 
  • #4
[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]
 
  • #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]
 

1. What is a convergent series?

A convergent series is a sequence of numbers that approaches a finite limit as the number of terms increases. In other words, the sum of the terms in the series will eventually become a fixed value as more terms are added.

2. What is a supremum norm?

A supremum norm is a way to measure the convergence of a series by finding the maximum distance between the series and a given function. It is also known as the "least upper bound" norm, as it is the smallest upper bound for the series.

3. What is an integral norm?

An integral norm is a method for measuring the convergence of a series by calculating the area under the curve of the series. It is also known as the "least integral bound" norm, as it is the smallest integral bound for the series.

4. How do supremum and integral norms help analyze convergent series?

Supremum and integral norms provide a way to determine the convergence of a series by comparing it to a known function or by calculating the area under the curve. These norms can help determine the behavior of a series and whether it approaches a finite limit or not.

5. What are some applications of analyzing convergent series with supremum and integral norms?

Analyzing convergent series with supremum and integral norms has many practical applications in various fields such as physics, engineering, and economics. It can be used to predict the behavior of complex systems, optimize processes, and solve differential equations. It is also a fundamental concept in mathematical analysis and plays a crucial role in understanding the convergence of numerical methods used in computer science and data analysis.

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