Originally Posted by

**SworD** Honestly, the comparison test is inadequate when it comes to gaining an understanding of these series. It is only good for formally proving that they converge/diverge, but the insight needs to come first.

The first series converges, but $\displaystyle \frac{n}{n^2+1}$ DIVERGES, so you cannot use this as a comparison. How do I know the series in #1 converges? Because the power in the numerator is 0.5, and the power of the denominator is 2.

If that doesn't make sense, I highly suggest that you become familiar with this general rule: a series in the form of a fraction will converge if the power of the denominator is higher than the power of the numerator by MORE than 1, and it will diverge if the power of the denominator exceeds the power of the numerator by less than 1, or if it doesn't exceed it at all. If the power of the denominator is higher than the power of the numerator by exactly 1, then the test cannot be used (but if both the numerator and denominator are polynomials, the series will diverge). By power, I always mean the highest power present. Something similar to this is called the polynomial test, but the polynomial test can actually be vastly generalized. Here are some examples:

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} $ converge, because (2 - 0) > 1

$\displaystyle \sum_{n=1}^{\infty} \frac{10000000n^3 + 500n^2}{n^{4.01} + n^2} $ converge, because (4.01 - 3) > 1

$\displaystyle \sum_{n=1}^{\infty} \frac{n^3}{10000000n^{3.999999}} $ diverge, because (3.999999 - 3) < 1

$\displaystyle \sum_{n=1}^{\infty} \frac{n^3 + 5n}{100n^4}$ diverge, because 4 - 3 = 1 and both are polynomials

$\displaystyle \sum_{n=1}^{\infty} \frac{n^3}{n^2 + 100000000n} $ diverge, because power in the numerator is greater

Acknowledging that any natural logs of polynomial functions of n have power 0, you can even determine series like these just by looking at them:

$\displaystyle \sum_{n=1}^{\infty} \frac{\ln{n^{9999}}}{n^2} $ converge, because power in the numerator is 0 and power of denominator is 2

However, using this method you could mess up if you don't know what you are doing, so the key is just to know how to use it and how it works.

Equipped with this understanding, you can easily generate a comparison for each of the problems that you need. For example, for #1, you can compare it to

$\displaystyle \frac{\sqrt{n}}{n^2+1} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{1.5}}$ converges (formally, an integral test can be used to prove that this p-series converges)

For #2, the numerator can be any upper bound constant:

$\displaystyle \frac{2+cos(n)}{n^2} < \frac{4}{n^2}$

For #3, the question is of a different nature, because it doesn't even pass the nth term test... the terms don't even go to 0, because dividing by 0.5^n makes the terms unbounded. So you can compare it with a constant:

$\displaystyle \frac{arcsec(n)}{0.5^{n}} > 1$ for sufficiently large n

so it has to diverge.