# Series Convergence/Divergence Questions

malindenmoyer
Determine if the following converge or diverge and explain why.

1. $$\Sigma_{n=1}^{\infty} \frac{\sin{\frac{n\pi}{2}}}{n}$$

2. $$\Sigma_{n=1}^{\infty} \frac{\ln{n}}{n^2}$$

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:

$$\frac{\ln{n}}{n^2} < \frac{n}{n^2} = \frac{1}{n}$$

and then

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

Is this the proper way to solve (2)?

Homework Helper
For the first one, write out the first 5 or 6 terms of the series numerically and see if that gives you any ideas. For the second one, no, bounding a series above by a divergent series doesn't prove it diverges. 1/n^2<1/n also, but 1/n^2 converges. Can you think of an upper bound comparison that converges?

malindenmoyer
Okay I got (2), (thanks for the correction), by comparing (2) to $$\frac{\sqrt{n}}{n^2}=\frac{1}{n^\frac{3}{2}}$$ which converges (p-series w/ p>1), therefore (2) converges. Is this correct?

As for (1) I wrote out the terms and discovered it is an alternating series with the even terms all equal to 0 as follows:

1-1/3+1/5-1/7+...

I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.

Homework Helper
Okay I got (2), (thanks for the correction), by comparing (2) to $$\frac{\sqrt{n}}{n^2}=\frac{1}{n^\frac{3}{2}}$$ which converges (p-series w/ p>1), therefore (2) converges. Is this correct?

As for (1) I wrote out the terms and discovered it is an alternating series with the even terms all equal to 0 as follows:

1-1/3+1/5-1/7+...

I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.

Yes, that's right for (2). For (1) you have an alternating series with decreasing terms. How about the alternating series test?

malindenmoyer
Yea I figured it had to do with that.

If we define the alternating series to be: $$\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n$$

then I can't seem to determine a $$b_n$$ which defines the series for all n, especially since every odd $$n$$ changes sign and all the even $$n$$'s are 0...I must be missing something...

If we define the alternating series to be: $$\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n$$
then I can't seem to determine a $$b_n$$ which defines the series for all n, especially since every odd $$n$$ changes sign and all the even $$n$$'s are 0...I must be missing something...