Radius of Convergence for Moderately Complicated Series

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Homework Help Overview

The discussion revolves around determining the radius of convergence for a series defined as \(\Sigma^{n=\infty}_{n=1}\left( \frac{n+p}{n}\right)^{n^{2}} z^{n}\). The original poster attempts to show that this radius is equal to \(exp\left(-p\right)\) and has made an initial attempt involving limits and logarithmic manipulation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the limit definition of the exponential function and its relation to the problem. There are inquiries about specific limits, particularly \(\lim_{n\to\infty}\left(1+\frac{p}{n}\right)^n\), which may be relevant to the original poster's approach.

Discussion Status

The discussion is ongoing, with participants providing clarifications and exploring different aspects of the problem. Some guidance has been offered regarding the limit definitions, but no consensus or resolution has been reached yet.

Contextual Notes

There may be assumptions regarding the properties of limits and the exponential function that are being examined, as well as the implications of the series' structure on its convergence.

LukeMiller86
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1. The problem statement:

Show that the following series has a radius of convergence equal to [tex]exp\left(-p\right)[/tex]

Homework Equations



For p real:

[tex]\Sigma^{n=\infty}_{n=1}\left( \frac{n+p}{n}\right)^{n^{2}} z^{n}[/tex]

The Attempt at a Solution


[tex]\stackrel{lim}{n\rightarrow\infty}\left|a_{n}\right|^{1/n} = \frac{1}{R} = \left(\frac{n+p}{n}\right)^{n}<br /> =exp\left(n\left(ln\left(\frac{n+p}{n}\right)\right)\right)[/tex]

Apart from playing with the logarithm after that I cannot seem to reach the required answer.
Any help would be greatly appreciated.
 
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What's the limit definition of the exponential function?
 
[tex]exp\left(-p\right) = e^{\left(-p)\right}[/tex]

is that what you meant?
 
Do you know this limit:
[tex]\lim_{n\to\infty}\left(1+\frac{p}{n}\right)^n[/tex]
 
Completely overlooked that! Thanks very much.
 

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