Testing Convergence: Comparison & Limit Comparison Tests

[ProBasket]

Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test.

$$\sum_{n=1}^\infty \frac{2n^4}{n^5+7}$$ this diverges using the p-series and comparison test right? p <1


$$\sum_{n=1}^\infty \frac{2n^4}{n^9+7}$$ and this converges right? because p > 1



$$\sum_{n=1}^\infty \frac{-1^n}{9n}$$ i think this also diverges cause p <1. (not sure about this one)

can someone check these real quick and tell me if I am correct?

 

[whozum]

First one, P = 1, it diverges.

2 is correct.

Hint for the last one

$$\sum_{n=1}^{\infty} - \frac{1^n}{9n}$$

Whats $1^n$ for positive n? It should be easy after that.

 

[HallsofIvy]

I suspect he meant $$\sum_{n=1}^{\infty} -\frac{(-1)^n}{9n}$$ for the last one. That converges because it is an alternating series with terms going to 0.

 

[Data]

I was going to make that suggestion, but since his post mentions the comparison test, I decided that the way he originally posted was likely the way that he meant it (since the comparison test can't be used for conditionally convergent series).