Show that $\lim_{n\to\infty} \sum^{\infty}_{k=n} \frac{n}{k^3}=0$

Question

Show that

$$\lim_{n\to\infty} \sum\limits^{\infty}_{k=n} \frac{n}{k^3}=0$$

This is a question related to measure theory.

Answer 1

Reformulation:

Let $\mu$ be the counting measure on $\Bbb N$. Then with: $$f_n(k)=\frac{n}{k^3} \chi_{\{k \geq n\}}$$ you are searching: $$\lim_{n \to +\infty} \int_{\Bbb N} f_n(k) d\mu(k)$$

Hint:

You can notice that:

• For all $k$: $$\lim_{n \to \infty} f_n(k)=0$$
• With $g(k)=\frac{1}{k^2}$, for all $n$: $$\forall k, \, |f_n(k)| \leq g(k)$$ and $$\lim_{n \to +\infty} \int_{\Bbb N} g(k) d\mu(k)<+\infty$$

Answer 2

We have that (see here) $$ \sum_{k=n}^\infty\frac1{k^3}\le\int_{n-1}^\infty x^{-3}dx=\frac12\cdot\frac1{(n-1)^2}. $$

Answer 3

For $k \geq n$, we have $$ \frac{n}{k^3} \leq \frac{1}{k^2} \ .$$ It follows that $$ 0 \leq \sum_{k = n}^\infty \frac{n}{k^3} \leq \sum_{k=n}^\infty \frac{1}{k^2} \ .$$ Since $$ \sum_{k=1}^\infty \frac{1}{k^2} < \infty \ ,$$ we must have $$ \lim_{n \to \infty} \sum_{k=n}^\infty \frac{1}{k^2} = 0 \ .$$ The result follows.