Series Convergence: Integral Test and Homework Equations

[razored]

Homework Statement

Determine whether this series converges or diverges :
http://mathbin.net/equations/7548_0.png

Homework Equations

Integral Test

The Attempt at a Solution

http://mathbin.net/equations/7548_1.png from that it shows to diverge but book says it converges. What did I do wrong?

 

[CompuChip]

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

 

[razored]

I don't understand what you mean.

 

[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!

 

[razored]

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

 

[CompuChip]

In general there are some rules which can help you determine whether your answer is correct. For example:

• 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.

 

[razored]

In general there are some rules which can help you determine whether your answer is correct. For example:

• 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.

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

 

[CompuChip]

Just to stress my point, they will not always help you.
First of all, you need to take into account that the sum will get an extra n (for example, if you can bound all terms in the sum from above by M, your sum will be like M n). For example: even though 1/n converges to zero, the sum over n = 1 to infinity of 1/n will go like n / n = 1 and will diverge.
Even then it is not guaranteed, for example, it is not clear if the sum over 1/(n log n) converges (for n > 1 for example; even if it does, it does so very slowly) although n/(n log n) goes to zero.
So you shouldn't use my general rules as more than rules of thumb, and always use a rigorous method to prove convergence.