Use of ratio test and comparison test to determine converging series Which of the following series converge, use ratio or comparison test to demonstrate: 


*

*$\sum_{n=1}^{\infty} \dfrac{n}{n^2+\cos^2(n)}$

*$\sum_{n=1}^{\infty} \dfrac{(-10)^n}{4^{2n+1}(n+1)}$

*$\sum_{n=1}^{\infty} \dfrac{n!}{3^n}$

*$\sum_{n=1}^{\infty} \dfrac{n}{(n+3)^3}$
Help much appreciated! Thanks!
 A: HINTS


*

*Note that $\cos^2(x)$ is always between $0$ and $1$. That sounds like a good thing to compare against.

*Just do it. There's a negative, so you might use the "alternating series test." It looks sort of geometric, so you might use the ratio test. Of the basic divergence test.

*A lesson for the future - factorials are hard to play with in limits and in series. The ratio test works nicely with factorials, thought (why?) so in general, you might want to associate the ratio test with factorials.

*Just do it. There's no need to do anything slick. Or alternately note that $n+3 > n$.

A: A slightly different set of HINTS:


*

*$n^2+\cos^2n\le2n^2$, so how is $\dfrac{n}{n^2+\cos^2n}$ related to $\dfrac{n}{2n^2}$?

*The ratio test often works well when you have products or quotients of exponentials with the index of summation in the exponent.

*As mixedmath said, the ratio test also behaves well with factorials of linear functions of the index of summation.

*$n<n+3$, so how is $\dfrac{n}{(n+3)^3}$ related to $\dfrac{n}{n^3}$? Alternatively, $$\sum_{n=1}^\infty\frac{n}{(n+3)^3}=\sum_{n=4}^\infty\frac{n-3}{n^3}<\dots\;?$$
