TransWikia.com

Why is the function `SumConvergence` not valid for simple power series

Mathematica Asked on January 10, 2021

I want to find the convergence interval of function $sum_{n=1}^{infty} frac{1}{3^{n}+(-2)^{n}} frac{x^{n}}{n}$.

Sum[(1/(3^n + (-2)^n))*(x^n/n), {n, 1, Infinity}]
SumConvergence[(1/(3^n + (-2)^n))*(x^n/n), n]

The above code can not get the convergence interval of the series $sum_{n=1}^{infty} frac{1}{3^{n}+(-2)^{n}} frac{x^{n}}{n}$, but we can easily calculate the convergence radius of the series $sum_{n=1}^{infty} frac{1}{3^{n}+(-2)^{n}} frac{x^{n}}{n}$ by using the convergence checking method:

Limit[1/(3^(n + 1) + (-2)^(n + 1))/(1/(3^n + (-2)^n)), n -> Infinity]

I want to know why the function SumConvergence is not valid for a simple power series $sum_{n=1}^{infty} frac{1}{3^{n}+(-2)^{n}} frac{x^{n}}{n}$.

Under the prompt of Michael E2, I can judge the convergence at the end point by ratio discrimination:

Limit[((-3)^n/(3^n + (-2)^n) 1/n)/((-1)^n/n), n -> Infinity]
Limit[n (3^n/(3^n + (-2)^n) 1/n), n -> Infinity]

One Answer

Try

SumConvergence[(1/(3^n + (-2)^n))*(x^n/n), n, Method -> "RatioTest"]
Abs[x] < 3

I found it by Reading The _ Manual, actually.

Correct answer by Michael E2 on January 10, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP