# Series - Testing for Convergence / Divergence

1. ### steelphantom

159
I have a few series which I'm having trouble proving whether they converge or diverge. I know the following tests for convergence: comparison test, ratio test, n-th term test, and root test. Here are the series and what I have tried so far:

$$\sum$$ n -1 / n2 : I'm assuming this series diverges, since it behaves like 1/n, which also diverges. I'm trying to use the comparison test to see if I can find a "smaller" series which also diverges, but coming up blank. I tried the ratio test to no avail, since it gives 1.

$$\sum_{n=2}^\infty$$ 1 / (n + (-1)n)2 : I'm really not sure where to begin with this one. The (-1)n is really throwing me off. I'm assuming this converges.

And finally,

$$\sum$$ n! / nn : I tried the ratio test, canceling out the factorial and getting the ratio of nn / (n + 1)n. This limit seems to be 1, so the ratio test doesn't really help me here. Any suggestions?

Thanks for any help!

Last edited: Feb 28, 2008
2. ### Dick

25,913
For the first one, how about saying (n-1)/n^2>=(n/2)/n^2? Notice that doesn't work for n=1, but that's ok. You can always ignore any finite number of terms. For the second one, your series is LARGER than 1/(2n)^2. That's no good. Being larger than a convergent series doesn't tell you much. How about 1/(n/2)^2? The last one is more delicate. Do you know Stirling's formula?

3. ### jhicks

335
For the third one, n^n/(n+1)^n does NOT approach 1 as n becomes large.

4. ### steelphantom

159
Ok, thanks for the help on the first two! I noticed that my series was larger than 1/(2n)^2 right after I posted. Dumb mistake. I don't know Stirling's formula. Is there another way to go about this?

jhicks, if n^n/(n+1)^n doesn't approach 1, I can only assume it approaches a number less than 1, in which case the series would converge. How could I show this limit is less than 1?

5. ### Dick

25,913
jhicks is right. (n/(n+1))^n=1/((n+1)/n)^n=1/(1+1/n)^n. Does that look familiar?

6. ### steelphantom

159
Ah, I see it now. Thanks a lot!