# Infinite Series & Convergence

kingwinner

## Homework Statement

Determine convergence for each of the following:

∑ 1 / [n (log n)2]
n=2

∑ 1 / [n log n log(log n)]
n=2

[log=ln=natural log]

## The Attempt at a Solution

I learned the root test, ratio test, comparsion test, and integral test. But how can we figure out what test to use? Would the "ratio test" work here?

Any help is appreciated!

Bohrok
I don't think the ratio test would help since you'll end up with ln(n + 1).
Try comparing n*(ln(n))2 with something that grows slower for x > 2.

kingwinner
hmm...Why wouldn't the ratio test work?? What's wrong with having ln(n+1)??

I'm just confused becuase I don't know which test to use when I look at those problems...

Should I use the integral test? Why or why not?

Thanks!

Bohrok
Yes, you can use the integral test (I should go to bed too tired)

There's no way to simplify ln(n + 1). I didn't look much at doing it with the ratio test; it might be possible, but I think it would be harder than any other way.

Staff Emeritus
Homework Helper
You just have to try each test out and see if it'll work out. Eventually, you'll develop a feel for which test is probably the best for a series.

kingwinner
I have some problem applying the integral test for the second sum...

Integral test: if f is continuous, decreasing, and POSITIVE on [2,∞), then

∑ f(k) converges if and only if
k=2

∫ f(x) dx converges.
2

f(x) = 1 / [x log x log(log x)]
But f is not positive on [2,∞)
e.g. f(2)<0, f(2.1)<0

So how can we apply the integral test??

Homework Helper
Ok, then apply it on [3,infinity).

kingwinner
Ok, then apply it on [3,infinity).

But is that ok? Will it affect convergence/divergence?

∫ f(x) dx
k
Does the value of "k" here have any effect on convergence/divergence?

Homework Helper
But is that ok? Will it affect convergence/divergence?

∫ f(x) dx
k
Does the value of "k" here have any effect on convergence/divergence?

You can drop any finite number of terms without affecting convergence or divergence of the series. So no, k doesn't really matter.

kingwinner
You can drop any finite number of terms without affecting convergence or divergence of the series. So no, k doesn't really matter.

For infinite series, the place where it starts has no effect on convergence/divergence.
But is it also true for improper INTEGRALS?

∫ f(x) dx converges
8

∫ f(x) dx diverges
5
Is this possible??

Homework Helper
For infinite series, the place where it starts has no effect on convergence/divergence.
But is it also true for improper INTEGRALS?

∫ f(x) dx converges
8

∫ f(x) dx diverges
5
Is this possible??

Sure, it's possible. That's why you have to check the premises of the integral test before you apply it. f(x) has to be nonnegative and monotone decreasing on the part of the series you want to apply it to. Just like in this case you should move k from 2 to 3.

kingwinner
Integral test: if f is continuous(?), decreasing, and POSITIVE on [n,∞), then

∑ f(k) converges if and only if
k=n

∫ f(x) dx converges.
n

To apply the integral test, does f have to be continuous? In my first year calculus text book, the integral test is as stated above; f is assumed to be continuous. But in my other textbook, there is no such assumption. Or maybe there is a typo?

What is the correct statement of the integral test? Do we need to assume continuity?

Gold Member
Yes, f(x) has to be continuous. If there is a infinite discontinuity, it will seriously affect convergence. If there is a jump discontinuity, then your function isn't likely to be easily integrable.

Of course, $$\frac{1}{x log(x) log(log(x))}$$ is quite easily integrable and is continuous from 3 to infinity. Go for it.

Gold Member
I believe (and I might be wrong) that an integral implies continuity.

Maybe even right... have you tried integrating 1/x over an interval that includes 0?

kingwinner
But I think it is continuity implies integrable, not vice versa...

What can happen to the "Integral Test" if f is not continuous???

(actually in my real analysis book, they didn't mention "continuity" in the integral test at all, weird...)