# Math Help - Infinite Series

1. ## Infinite Series

I wrote it in word instead :P

2. Solved second part just need help with first part now!=)

3. Still need some help on where to start. Thanks

4. Originally Posted by Len
Would also be much appreciated if someone can confirm my use of the comparison test

Forgot in the first line for all positive integers k.
Your use of the comparison test is fine:

$\frac{\sqrt{k} 4^k}{2 + k 5^k} < \frac{k 4^k}{k 5^k} = \frac{4^k}{5^k} = \left( \frac{4}{5} \right)^k$.

Without numerical number crunching, I'm not sure how to establish how large n should be.

5. Originally Posted by mr fantastic
Your use of the comparison test is fine:

$\frac{\sqrt{k} 4^k}{2 + k 5^k} < \frac{k 4^k}{k 5^k} = \frac{4^k}{5^k} = \left( \frac{4}{5} \right)^k$.

Without numerical number crunching, I'm not sure how to establish how large n should be.
Thanks

Maybe someone else can help me with the first part.