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]
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
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