# Power Series Convergence intervlas Hi! I am working on some online calc2 homework problems on power series and I am completely confused on how to solve these types of questions. I really do not know how to begin to tackle this problem. If someone has a free minute to help me out by walking me through this problem step by step I would greatly appreciate it!

When dealing with power series, the ratio test is often the right test for convergence (the root test can be used if you have terms like $n^n$ or other exponentials, but more often than not, these can still be dealt with using the ratio test). So if we use the ratio test, we get: \begin{align*} \lim_{n\to\infty}\left|\frac{ \frac{(x+1)^{n+1}}{(n+1)^4 + 10}}{\frac{(x+1)^n}{n^4+10}} \right| &= \lim_{n\to\infty}\left|\frac{(x+1)^{n+1}}{(n+1)^4 + 10} \cdot \frac{n^4+10}{(x+1)^n}\right|\\ &= \left| (x+1) \right|\lim_{n\to\infty} \frac{n^4+10}{(n+1)^4+10}\\ &= \left|(x+1)\right| \lim_{n\to\infty} \frac{n^4(1+\frac{10}{n^4})}{n^4((1+\frac{1}{n})^4 + \frac{10}{n^4})}\\ &= \left|(x+1)\right| \lim_{n\to\infty} \frac{(1+\frac{10}{n^4})}{((1+\frac{1}{n})^4 + \frac{10}{n^4})}\\ &= \left|(x+1)\right| < 1 \end{align*} which is true if and only if $-1 < x+1 < 1$, or $-2 < x < 0$
If $x = -2$, the series is \begin{align*} \sum_{n=0}^\infty \frac{(-1)^n}{n^4+10} < \infty \end{align*} which converges absolutely by the limit comparison test (compare with $\frac{1}{n^4}$). Also, if $x = 0$, the series is \begin{align*} \sum_{n=0}^\infty \frac{1}{n^4+10} < \infty \end{align*} which also converges absolutely by the limit comparison test (again comparing with $\frac{1}{n^4}$).
So the interval of convergence is then $[-2, 0]$