# Series - Integral Test

1. Apr 4, 2009

### razored

1. The problem statement, all variables and given/known data
Determine whether this series converges or diverges :
http://mathbin.net/equations/7548_0.png [Broken]

2. Relevant equations
Integral Test

3. The attempt at a solution
http://mathbin.net/equations/7548_1.png [Broken] from that it shows to diverge but book says it converges. What did I do wrong?

Last edited by a moderator: May 4, 2017
2. Apr 4, 2009

### CompuChip

Where did you get the ln| ... | from?
Try $1/e^{n^2}$ as a primitive.

3. Apr 4, 2009

### razored

I don't understand what you mean.

4. Apr 4, 2009

### CJ2116

Just use a substitution for n2:

Let u = n2 and du/2n = dn. The n's will cancel leaving you with an easy integral to evaluate. Then use the fundamental theorem to get a (convergent) answer. There shouldn't be any Ln's in your answer!

5. Apr 4, 2009

### razored

Oh, so the integral is $$\frac{-1e^{-n^2}}{2}$$. Okay, now it converges. Thanks!

6. Apr 5, 2009

### CompuChip

• in polynomials, the highest power wins. For example, x/x^2 converges, $6 x^3 \sqrt{x} / (12 x^4)^{5/6}$ diverges because it goes like $x^{7/2}/x^{10/3} = x^{1/6}$ which has positive power.
• logarithms converge slower than polynomials, e.g. x/log(x) will diverge.
• exponentials go faster than any polynomial, for example $x^{999999}/e^x$ converges while $x^{-9999999} e^x$ diverges.

You will find more of these as you go, they can often help you make an estimate.
In this case, the terms in the sum converge to 0 much faster than 1/n, so in general you expect a convergent sum.

7. Apr 5, 2009

### razored

My text book left those out. They probably assumed the student would figure them out themselves. Thank you for pointing them out!

8. Apr 5, 2009