# Thread: Calculus III - trouble with the comparison test

1. ## Calculus III - trouble with the comparison test

I am working on a few problems at home to get the hang of the comparison test but I'm having trouble finding what to compare them to. I understand the theorem and I understand what is being asked, I just draw a blank when I face a new problem. Any tips, suggestions, or shortcuts? These are the problems I'm working on:
$1.) \sum_{n=1}^{\infty }\frac{\sqrt[]{n}}{n^2+1}$
For this one I thought comparing it to $\frac{n}{n^2+1}$ would show it converges since it is greater than and it converges. Is that a logical jump or is there a better answer?

$2.) \sum_{n=1}^{\infty }\frac{2+cos(n)}{n^2}$

$3.) \sum_{n=1}^{\infty }\frac{arcsec(n)}{0.5^n}$

On the last two I drew a complete blank. I've been thinking what to compare them to but all of the ones I think of are less than the problem. It's so infuriating!

Thank you so much for your help.

2. ## Re: Calculus III - trouble with the comparison test

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 $\frac{n}{n^2+1}$ 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:

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

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

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

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

$\sum_{n=1}^{\infty} \frac{n^3}{n^2 + 100000000n}$ 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:

$\sum_{n=1}^{\infty} \frac{\ln{n^{9999}}}{n^2}$ 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
$\frac{\sqrt{n}}{n^2+1} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{1.5}}$ 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:

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

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:

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

so it has to diverge.

3. ## Re: Calculus III - trouble with the comparison test

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 $\frac{n}{n^2+1}$ 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:

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

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

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

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

$\sum_{n=1}^{\infty} \frac{n^3}{n^2 + 100000000n}$ 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:

$\sum_{n=1}^{\infty} \frac{\ln{n^{9999}}}{n^2}$ 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
$\frac{\sqrt{n}}{n^2+1} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{1.5}}$ 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:

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

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:

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

so it has to diverge.
SworD, you are my knight in shinning armor. Thank you so much! This makes so much sense I wish I had understood this from the beginning. I will be printing out your reply and adding it to my notes. Although my homework asks specifically for me to prove with the comparison test, I will ask my professor if I can do it another way. Thanks again!

4. ## Re: Calculus III - trouble with the comparison test

No problem, if there's a part of my post that you didn't understand or you have any questions feel free to post them or send me a PM.

5. ## Re: Calculus III - trouble with the comparison test

Another thing that might be useful to note, that I didn't mention: exponential functions where the base > 1 have infinite "power". Factorials have infinite power as well. So series like this can automatically be determined to converge:

$\sum_{n=1}^{\infty} \frac{n^2 + 50n + 10}{e^n}$ converge
$\sum_{n=1}^{\infty} \frac{n^{9999} + 80n^2}{1.00001^n}$ converge
$\sum_{n=1}^{\infty} \frac{3n^2 + \sqrt{n}}{2^n}$ converge
$\sum_{n=1}^{\infty} \frac{10n^{100} + 3n^8}{n!}$ converge

If the base < 1, the "power" is negative infinity. So these trivially diverge (the power of the numerator is some number, but the "power" of the denominator is negative-infinite. So in a manner of speaking, the power of the numerator is greater, which clearly causes the series to diverge.)

$\sum_{n=1}^{\infty} \frac{2n^2 + 30n + 10}{0.5^n}$ diverge
$\sum_{n=1}^{\infty} \frac{1}{0.9999^n}$ diverge

Furthermore, if you multiply two terms, you add their power. Think of normal power functions being the primary example. $x^2 \cdot x^3 = x^5$. You multiplied the terms, the powers were added. So, even series which look big and bad, like this one can be determined:

$\sum_{n=1}^{\infty} \frac{n^2 + n^5\ln(n^2 + 5n)}{n^7\ln(n) + \ln(2n)}$ Converges. The term with the highest power in the numerator is $n^5\ln(n^2 + 5n)$: you add the power of the n^5 to the power of the ln (which has 0 power) to get 5. Similarly in the denominator, you'll get 7.

A whole class of the standard converge/diverge problems can be solved on sight if you have a good understanding of this. Some will require more work, though. Be careful that you recognize these and don't jump the gun. For example,

$\sum_{n=1}^{\infty} \frac{\tan(n)}{n^2}$ It is difficult to assign a "power" to tan(n), so the test is meaningless here, unless further work is done to establish some kind of limiting behavior.

$\sum_{n=1}^{\infty} \frac{n^n}{e^n n!}$ Both the numerator and the denominator have infinite "power", so any comparison is meaningless as well. This divergence of this particular series is actually extremely difficult to establish.

$\sum_{n=1}^{\infty} \frac{n^2 + n^6\ln(n^2 + 5n)}{n^7\ln(n) + \ln(2n)}$ Very similar to the example above, but the powers are now 6 and 7: since they differ by precisely 1, the test is silent.

Also keep in mind that I have used a somewhat sketchy/informal notion of "power". It can be proven rigorously that to determine convergence/divergence of series it is okay to extend the notion of power like this, but the term "power", formally, is usually only applied to functions that actually have the form x^p.