# Does $\sum_{n=1}^\infty \frac{n+2}{n^2+4n-7}\left(\frac{4}{7}\right)^n$ converge?

Does the following series converge, does it converge absolutely? $$\sum_{n=1}^\infty \frac{n+2}{n^2+4n-7}\left(\frac{4}{7}\right)^n$$

Now I thought about using Dirichlets Test of Uniform Convergence.

I know that $|\sum_{n=1}^\infty (\frac{4}{7})^n|\leq M < \infty$.

I am left to show that if $\frac{n+2}{n^2+4n-7} \rightarrow 0$ uniformly then my whole series converge uniformly?

What does uniform convergence of series mean?

I know that it means partial summs converge uniformly, but what is it in lazy mans terms? Does it imply absolute convergence?

• This problem will succumb to a simple Comparison Test. Give it a try. – hardmath May 8 '16 at 18:07
• It will be easier the other way around, $\forall n \geq 2, \frac{n+2}{n^2+4n-7} \leq 1$. Then use the convergence of $\sum_{n = 1}^N (\frac{4}{7})^n$ – Vincent May 8 '16 at 18:12
• Root test is also good. – Slowpoke May 8 '16 at 18:13

Hint. One may just observe that $$0\leq \sum_{n=2}^\infty \frac{n+2}{n^2+4n-7}\left(\frac{4}{7}\right)^n\leq \sum_{n=2}^\infty \left(\frac{4}{7}\right)^n<\infty.$$
hint: $\dfrac{n+2}{n^2+4n-7} < 1 (n \geq 3)$ and $\left|\dfrac{4}{7}\right| < 1$