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The problem is to find the radius of convergence and the interval of convergence for the following power series:

$$\sum_{n=0}^{\infty}\frac{x^n}{3^n(n^2+1)}$$

My Attempt:

I know that the radius of convergence for a power series $$\sum_{n=0}^{\infty}c_n(x-a)^n$$ is given by $$R = \frac{1}{\limsup_{n\to\infty}|c_n|^{1/n}}$$

So, I tried to apply this formula to the given series, treating $$c_n = \frac{1}{3^n(n^2+1)}$$ and $$a = 0$$.

However, I'm not sure how to calculate the limit superior of the sequence $$|c_n|^{1/n} = \left|\frac{1}{3^n(n^2+1)}\right|^{1/n}$$

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  • $\begingroup$ In this particular case you may use the d'Alembert test $${|c_n|\over |c_{n+1}|}=3{(n+1)^2+1\over n^2+1}\to 3$$ $\endgroup$
    – Ryszard Szwarc
    Commented May 6 at 19:06

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You are almost done, indeed we have

$$ \left|\frac{1}{3^n(n^2+1)}\right|^{1/n}= \frac13 \frac{1}{(n^2+1)^{1/n}}$$

with $(n^2+1)^{1/n}\to 1$.

Finally we need to discuss separately the cases $x=\pm 3$.

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