Does anybody see how to tell if this serie converges.(adsbygoogle = window.adsbygoogle || []).push({});

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

where log is the neperian logarithm.

This is my latest attempt...

Cauchy's [itex]n^{th}[/itex] root criterion:

[tex]\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}[/tex]

But

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

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

[tex]\left(\frac{n+4}{2n}\right)^{logn} < \left(\frac{n+4}{2n}\right)^{loge} < \frac{n+4}{2n}[/tex]

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

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# Convergence of a serie

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