Mutaja said:
How do I proceed then, if my comparison test is inconclusive? Let's leave my b) problem out of this for now. For my a) problem, jbunniii was kind enough to suggest to try the comparison test. My result from the comparison test is that it's inconclusive since $$\lim_{n => ∞}\frac{1}{2n}$$ approaches 0.
I don't think you have used the comparison test correctly here. Suppose we wish to test the convergence of ##\sum_n a_n##. For convenience we'll assume that ##a_n \geq 0## for all ##n## as is the case in this problem. Suppose I can find another series ##\sum_{n}b_n## (with ##b_n \geq 0##), which is known to diverge, and suppose I can establish that ##a_n \geq b_n## for all ##n## (or, more generally, for all sufficiently large ##n##). Then it follows that ##\sum_{n} a_n## must diverge.
Roughly speaking, ##\sum_n b_n = \infty## and ##\sum_n a_n## is "bigger" so it must also be ##\infty##.
So apply this to your problem: note that ##2n^2 - 1 < 2n^2## for all ##n##, so
$$\frac{1}{2n^2 - 1} > \frac{1}{2n^2}$$
for all ##n##. Multiplying both sides by the positive number ##n##, we get
$$\frac{n}{2n^2 - 1} > \frac{n}{2n^2} = \frac{1}{2n}$$
Thus *if* we know that ##\sum_{n} 1/(2n)## diverges, then ##\sum_{n} n/(2n^2 - 1)## must also diverge.
Reading your later posts, I can see that you didn't realize that
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
diverges. The terms get smaller and smaller, but they don't get small fast enough for the sum to converge to a finite number. The proof of this is quite simple. Write out the first few terms of the sum:
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \cdots$$
Let's group the terms together. The first group will have 1 term, the second group will have 2 terms, the third group will have 4 terms, and in general, the n'th group will have ##2^{n-1}## terms:
$$1 + \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right) + \cdots$$
Now we'll apply the comparison test as follows:
$$1 + \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right) + \cdots > 1 + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots$$
On the right hand side, each group sums to ##1/2##, and we are adding infinitely many of them. Thus the right hand side is ##\infty##, and since the partial sums of the left side are even bigger, the sum is also ##\infty##. So this proves that
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
diverges. This is a good piece of information to commit to memory because it will be useful in comparison tests and other contexts. It's called the harmonic series, by the way:
http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)
One interesting thing is that the harmonic series "barely" diverges. It crawls very slowly on its way to infinity. You have to add the first ##10^{43}## terms to even get to 100. Also, if instead of summing ##\sum_{n} 1/n##, you sum ##\sum_{n} 1/n^p## where ##p## is any number greater than 1 (for example, 1.000000000000000001), the series converges!
But I do like that approach too, because it shows why the sum diverges so slowly: the partial sums grow logarithmically. And also, comparing with integrals allows one to establish the convergence of ##\sum 1/n^p## for ##p>1##, which is not so easy to do directly.
