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.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Convergence of a serie

Loading...

Similar Threads - Convergence serie | Date |
---|---|

I Divergence/Convergence for Telescoping series | Mar 25, 2017 |

I Convergence of Taylor series in a point implies analyticity | Dec 10, 2016 |

A Convergence of an infinite series of exponentials | Nov 2, 2016 |

I Alternating Series, Testing for Convergence | Apr 13, 2016 |

Converging series | Jan 17, 2016 |

**Physics Forums - The Fusion of Science and Community**