# Math Help - Another Series: Looking at conv vs. div

1. ## Another Series: Looking at conv vs. div

So we have summation of 1/[k(k+1)(k+2)]

We know that 1/[k(k+1)(k+2)] < 1/[(k+1)(k+2)] which I know converges.

Thus, 1/[k(k+1)(k+2)] also converges.

Can someone check my reasoning?

2. Originally Posted by WartonMorton
So we have summation of 1/[k(k+1)(k+2)]

We know that 1/[k(k+1)(k+2)] < 1/[(k+1)(k+2)] which I know converges.

Thus, 1/[k(k+1)(k+2)] also converges.

Can someone check my reasoning?

I think this is fine. (Just make sure you have the proof that 1/[(k+1)(k+2)] converges).
As an exercise you can try to formally prove it using your argument above and the definition of convergence. It will give you good experience.

(Also you may want to put these/related questions in the analysis sub-forum.)