Convergent Series

[BSCowboy]

I have been thinking about this problem:
Determine whether the following series are convergent in $$\left(C[0,1],||\cdot ||_{\infty}\right)$$ and $$\left(C[0,1],||\cdot ||_{1}\right)$$.
when
$$f_n(t)=\frac{t^n}{n}$$

In the supremum norm, this seems pretty straightforward, but in the integral norm I am confused since,
$$\left\|\sum\frac{t^n}{n}\right\|_1\leq\sum\left\|\frac{t^n}{n}\right\|_1=\sum\int_0^1\frac{t^n}{n}dt=\sum\left[\frac{t^{n+1}}{n^2+n}\right]_0^1=\sum\frac{1}{n^2+n}<\sum\frac{1}{n^2}$$
and, I think this converges as $$n\rightarrow\infty$$, 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?

 

[mathman]

Could you be more precise as to what is supposed to be converging to what? Partial sums of f_n(t) or f_n(t) itself as a sequence.

 

[BSCowboy]

Actually, I think I might have figured it out:
$$f_n(t)=\frac{t^n}{n} \text{ in } \left(C[0,1],\|\cdot\|_1\right)$$

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

But, because the space $$\left(C[0,1],\|\cdot\|_1\right)$$ is not complete all we know is that $$f_n(t)$$ 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].

 

[statdad]

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

but

$$\lim_{n \to \infty} \frac 1 {n(n+1)} \to 0$$

 

[adriank]

Also $$\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left[ \frac{1}{n} - \frac{1}{n+1} \right] = 1$$