enter image description here

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 $-2 < x < 0$

We still need to check endpoints:

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 $[-2, 0]$

  • $\begingroup$ Thank you so much! I am still struggling with this topic, but your explanation helped to clear up a few things. $\endgroup$ – user124539 Apr 2 '14 at 3:10
  • $\begingroup$ Take some time to look this over, and feel free to ask if anything is unclear about the explanation, or the topic in general. Power series aren't an easy thing to learn, and it takes a bit of practice and a lot of time to get them down. $\endgroup$ – Nicholas Stull Apr 2 '14 at 3:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.