Show that the series diverges $\sum_{k=2}^\infty \frac{k^{k-2}}{3k}$

Question

Show that the series diverges $$\sum_{k=2}^\infty \frac{k^{k-2}}{3k}$$

This can be written as

$$\frac{1}{3}\sum_{k=2}^\infty {k^{k}}{k^{-3}}$$

How can I mathematically establish with the geometric series rule this going to diverge?

Answer 1

Hint: Prove that $$k^k\geq 3k^2$$ for $$k\geq k_0$$

Answer 2

We have that

$$\frac{k^{k-2}}{3k}=\frac{k^k}{3k^3}\to \infty$$

therefore the series diverges indeed recall that

$$\sum a_k <\infty \implies a_k \to 0$$

Answer 3

$k \ge 4$;

$k^{k-3} >1;$

$0<(1/3)<(1/3)k^{k-3}$;

Series diverges.

Answer 4

Use induction to show that $k^k \ge 3k^3$. The base step is $k=4$ and the induction step follows from

$$(k+1)^{k+1}=(k+1)^{k}(k+1)\ge 3(k+1)^k \ge 3(k+1)^3$$