Infinite Series

**Len** · April 2nd 2008, 11:28 AM

I wrote it in word instead :P

**Len** · April 2nd 2008, 12:31 PM

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

**Len** · April 2nd 2008, 02:19 PM

Still need some help on where to start. Thanks

**mr fantastic** · April 2nd 2008, 04:17 PM

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.

**Len** · April 2nd 2008, 04:24 PM

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.

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