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]

1 Answer 1



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

I found it by Reading The _ Manual, actually.

  • $\begingroup$ But the point of x = 3 is also convergent. However, the above results do not include the case of x = 3. $\endgroup$ Aug 22, 2020 at 2:56
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    $\begingroup$ @Montevideo The Ratio Test does not apply to endpoints, only to the radius of convergence. Also, it's a complex disk, so there are other points to check as well, I think. -- Maybe this is the problem: SumConvergence[(1/(3^n + (-2)^n))*(x^n/n) /. x -> 3, n] $\endgroup$
    – Michael E2
    Aug 22, 2020 at 2:58
  • $\begingroup$ SumConvergence[(1/(3^n + (-2)^n))*(x^n/n) /. x -> -3 I, n] doesn't work either. $\endgroup$
    – Michael E2
    Aug 22, 2020 at 3:04
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    $\begingroup$ @Montevideo That's the n-th term test, and all it shows is that it fails to show divergence. Multiplying the term by n gives an indeterminate result (limit = 1): Limit[n (1/(3^n + (-2)^n) 3^n/n), n -> Infinity]. $\endgroup$
    – Michael E2
    Aug 22, 2020 at 3:11
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    $\begingroup$ Multiplying by n^2 is too big. If the limit is finite, then good, it's convergent; otherwise it's indeterminate. Mutliplying by n^(1+e) for any e > 1 has the same issue. $\endgroup$
    – Michael E2
    Aug 22, 2020 at 3:29

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