Testing series for convergence

**Niall101** · October 8th 2008, 12:03 PM

Asked to test the following for convergence using the comparason test as much as possible.

1. 1/(n^p-n^q) 0<q<p I thin kthe limit in this one does not exist but cant say why.

2. sqrt(N^2+1) - n

3. (n^(1/n) -1) I think this converges to a limit of -1 but cant show why

Please help I am totally stuck.

Regards
Niall101

**watchmath** · October 8th 2008, 07:36 PM

Your notation is not so clear.
Do you mean for example number 2 is showing the convergence/divergence of $\sum \sqrt{n^2+1}-n$ ?

$\sqrt{n^2+1}-n=\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{(n+1)+n}>\frac {1}{2}\frac{1}{n+1}$
So the series is divergent.

**CaptainBlack** · October 8th 2008, 08:22 PM

Originally Posted by Niall101

1. 1/(n^p-n^q) 0<q<p I thin kthe limit in this one does not exist but cant say why.

$ \frac{1}{n^p-n^q}=\frac{1}{n^q(n^{p-q}-1)} $

but $p-q<0$ , so:

$ \lim_{n \to \infty} \frac{1}{n^{p-q}-1}=-1 $

and as $q>0$ :

$ \lim_{n \to \infty} \frac{1}{n^q}=0 $

So:

$ \lim_{n \to \infty}\frac{1}{n^p-n^q}=0 $

CB

**CaptainBlack** · October 8th 2008, 08:28 PM

Originally Posted by Niall101

2. sqrt(N^2+1) - n

$ \lim_{n \to \infty} \sqrt{n^2+1} - n = \lim_{n \to \infty} n(1+n^{-2})^{1/2}-n $

Now expand the square root as a power series to see what the limit is

CB

**watchmath** · October 8th 2008, 08:34 PM

The tittle seems that we are talking about the convergence/divergence of series here.

**CaptainBlack** · October 8th 2008, 08:34 PM

Originally Posted by Niall101

3. (n^(1/n) -1) I think this converges to a limit of -1 but cant show why

Not surprised as it goes to 0.

The -1 makes this more complicated than it need be so lose it:

$\lim_{n \to \infty} (n^{1/n} -1)=[\lim_{n \to \infty} n^{1/n}] -1$

Now you could find the limit on the right by looking at the limit of its log.

RonL

**CaptainBlack** · October 8th 2008, 08:37 PM

Originally Posted by watchmath

The tittle seems that we are talking about the convergence/divergence of series here.

If one finds the limits and they are finite then they converge. To find all the limits may not be what was intended but it is convincing and proves what was asked if not by the means intended.

It also simplifies the process of testing if you know the answer (that is if they do or do not converge)

Also, at a number of points in my posts one can just use tests to check for convergence rather than proceed to actualy find the limits.

CB

**watchmath** · October 8th 2008, 08:45 PM

Originally Posted by CaptainBlack

If one finds the limits and they are finite then they converge. To find all the limits may not be what was intended but it is convincing and proves what was asked if not by the means intended.

It also simplifies the process of testing if you know the answer (that is if they do or do not converge)

Also, at a number of points in my posts one can just use tests to check for convergence rather than proceed to actualy find the limits.

CB

I am not sure if you understand what I meant and I understand what you mean.
The title of the thread is about the convergence of series (NOT sequence) and also he/she talked about comparison test.

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