how to show the series diverges for the sum from k=2 to infinity k^(k-2)/3k

Question

how to show the series diverges for the sum from k=2 to infinity  k^(k-2)/3k

anonymous · Answer

$$\sum_{2}^{\infty} k ^{k-2}/3k$$

amistre64 · Answer

one thought is to do the integral test

anonymous · Answer

but k on its power
so i need to use ln?

anonymous · Answer

use the ratio test will be much easier

amistre64 · Answer

$$\sum_{2}^{\infty} k ^{k-\cancel{2}^3}/3\cancel{k}$$

anonymous · Answer

i havent learned the ratio test yet =[

amistre64 · Answer

1/3 pulls out and your left with 
$$1/3\sum_{2}^{\infty} k ^{k-3}$$

amistre64 · Answer

this should be self evident that it blows up to infinity

anonymous · Answer

it should be K^(k-2) not K^(K-2^3) tho

amistre64 · Answer

noone did k^k-2^3  thats a cancel and becomes a 3

amistre64 · Answer

we have 1/k = k^-1

k^k-2-1=k^k-3

anonymous · Answer

should i use ln to (K-3) down first

amistre64 · Answer

dunno, integral test was my first thought but i abandoned it for something a bit logical :)

amistre64 · Answer

hitting post is like playing the lotto, will it freeze or post? lol

anonymous · Answer

loll what do you mean

amistre64 · Answer

just trying to work the site.  i go to hit post to respond and my system will freeze up sometimes

anonymous · Answer

i don't have the problem, its probably because this site updates every second

amistre64 · Answer

but yeah, if you can show k^(k-3) does off into infinity then it diverges by definition

amistre64 · Answer

goes off into ...

amistre64 · Answer