Solve Series Convergence: Cauchy's n^{th} Root Criterion

[quasar987]

Does anybody see how to tell if this serie converges.

\sum \left(\frac{n+4}{2n+3} \right)^{nlogn}

where log is the neperian logarithm.

This is my latest attempt...

Cauchy's n^{th} root criterion:

\lim_{n \rightarrow \infty} \left( \left| \frac{n+4}{2n+3} \right|^{nlogn} \right)^{\frac{1}{n}} = \lim_{n \rightarrow \infty} \left(\frac{n+4}{2n+3}\right)^{logn}

But

\left(\frac{n+4}{2n+3}\right)^{logn} < \left(\frac{n+4}{2n}\right)^{logn}

And because for n \geq 5, (n+4)/2n &lt; 1 and because logx is a strictly increasing function, for n \geq 5, loge < logn and therefor

\left(\frac{n+4}{2n}\right)^{logn} &lt; \left(\frac{n+4}{2n}\right)^{loge} &lt; \frac{n+4}{2n}

But that's unsignificant; I would have to bound it superiorly with a sequence that goes to 0.

 

[shmoe]

when n is large, your terms "look like" (1/2)^{n\log n}, which is very convergent. This isn't at all precise, but gives some ideas on how to proceed. Your terms are actually slightly larger than this similar looking thing. So I'd try a comparison test with something like a^{n\log n} where a is a little bigger than 1/2.

 

[shmoe]

Sorry, you can ignore my last post and finish off your root test. You get to

\left(\frac{n+4}{2n}\right)^{\log n}

Your next inequality was no good-you bounded log(n) above by log(e).

Instead, try bounding \frac{n+4}{2n} above by something less than 1 and leaving the log(n) in the exponent alone. Remember it's fine if your bound only holds for large n.

 

[quasar987]

Ahhh, thanks shmoe!