Find the convergence interval of the power series $\sum_{n=1}^{\infty} \frac{x^n}{n^4}$

Key takeaway: For a power series, the radius usually comes from the ratio or root test, but the endpoints still need a separate check. This one is a nice example because the denominator is strong enough to make the endpoints behave well.

We study

$\sum_{n=1}^{\infty} \frac{x^n}{n^4}.$

Step 1: Find the radius of convergence

Use the ratio test on

$a_n=\frac{x^n}{n^4}.$

Then

=\left|\frac{x^{n+1}}{(n+1)^4}\cdot\frac{n^4}{x^n}\right| =|x|\left(\frac{n}{n+1}\right)^4.$$ Taking the limit gives $$\lim_{n\to\infty} |x|\left(\frac{n}{n+1}\right)^4=|x|.$$ So the series converges when $$|x|<1,$$ which means the radius of convergence is $$R=1.$$ ### Step 2: Check the endpoints - **At $x=1$**: $$\sum_{n=1}^{\infty} \frac{1}{n^4}$$ is a $p$-series with $p=4>1$, so it converges. - **At $x=-1$**: $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}$$ converges absolutely because $$\sum_{n=1}^{\infty} \left|\frac{(-1)^n}{n^4}\right|=\sum_{n=1}^{\infty} \frac{1}{n^4}$$ also converges. ### Final answer The interval of convergence is $$\boxed{[-1,1]}.$$ --- **Pitfalls the pros know** 👇 A common slip is stopping after the ratio test and forgetting the endpoints. For power series, that last check matters. Here both endpoints are fine, but you only know that after testing them separately. Another easy mistake is thinking $x=-1$ is automatically conditional convergence just because there is a $(-1)^n$ factor. In this problem it is actually absolute convergence. **What if the problem changes?** If the denominator were $n^p$ instead of $n^4$, the same ratio test would still give radius $R=1$. The endpoint behavior would depend on $p$: at $x=1$ and $x=-1$, the series would converge only when $p>1$. If $p\le 1$, the endpoints would fail. `Tags`: ratio test, power series, p-series