- #1

malindenmoyer

- 31

- 0

1. [tex]\Sigma_{n=1}^{\infty} \frac{\sin{\frac{n\pi}{2}}}{n}[/tex]

2. [tex]\Sigma_{n=1}^{\infty} \frac{\ln{n}}{n^2}[/tex]

For (1) I know that the numerator is bounded by -1 and 1 but do not know how to solve it from there.

For (2) I used the comparison test and did the following:

[tex]\frac{\ln{n}}{n^2} < \frac{n}{n^2} = \frac{1}{n}[/tex]

and then

[tex]\Sigma_{n=1}^{\infty} \frac{1}{n}[/tex] diverges (harmonic series) therefore (2) diverges by Comparison Test.

Is this the proper way to solve (2)?