So, the infinite series converges for a>2 and diverges for a=2.

[ShizukaSm]

Homework Statement

Show that the infinite series
$\sum_{n=0}^{\infty} (\sqrt{n^a+1}-\sqrt{n^a})$

Converges for a>2 and diverges for x =2

The Attempt at a Solution

I'm reviewing series, which I studied a certain time ago and picking some questions at random, I can't solve this one.

I tried every test I could think of, they were ALL inconclusive, I believe there's some function I must came up with to use with the comparison test, but I can't figure which one is it.

 

[LCKurtz]

Homework Statement

Show that the infinite series
$\sum_{n=0}^{\infty} (\sqrt{n^a+1}-\sqrt{n^a})$

Converges for a>2 and diverges for x =2

The Attempt at a Solution

I'm reviewing series, which I studied a certain time ago and picking some questions at random, I can't solve this one.

I tried every test I could think of, they were ALL inconclusive, I believe there's some function I must came up with to use with the comparison test, but I can't figure which one is it.

Did you try rationalizing the numerator first?

 

[ShizukaSm]

Ohh, no I didn't. I didn't even think about that, but now that you mentioned it, it worked perfectly, I could apply the comparison test in the rationalized numerator and find my answer, beautiful!

Thanks a LOT LCKurtz :D

Ps: I really wish I had more of a "math view" that you guys in this forum frequently have, and identify when I should rationalize, when I should expand fractions, etc.

 

[Mark44]

Homework Statement

Show that the infinite series
$\sum_{n=0}^{\infty} (\sqrt{n^a+1}-\sqrt{n^a})$

Converges for a>2 and diverges for x =2

Since x isn't present in this problem, I assume you mean a = 2, and possibly a ≤ 2.

The Attempt at a Solution

I'm reviewing series, which I studied a certain time ago and picking some questions at random, I can't solve this one.

I tried every test I could think of, they were ALL inconclusive, I believe there's some function I must came up with to use with the comparison test, but I can't figure which one is it.

 

[ShizukaSm]

Hey mark44!

Yes, I apologize, I actually meant a>2 and not x>2, and yes, it's bigger than. If you do the rationalization it will become clear that you can compare the a=2 (when it becomes divergent) with the harmonic series, and a>2 will have an exponent smaller than one (p-series with p <1, that is, convergent.)

 

[Mark44]

Hey mark44!

Yes, I apologize, I actually meant a>2 and not x>2, and yes, it's bigger than.

You wrote a > 2 for convergence, which was OK. The other part had to do with divergence of the series, and you wrote x = 2. My correction was to change this to a = 2.

If you do the rationalization it will become clear that you can compare the a=2 (when it becomes divergent) with the harmonic series, and a>2 will have an exponent smaller than one (p-series with p <1, that is, convergent.)

 

[ShizukaSm]

Ok so I thought I figured out with the rationalization tip, but I think I only solved half of the problem with it:

$\frac{1}{\sqrt{n^a +1}+\sqrt{n^a}} < \frac{1}{\sqrt{n^a}}$ For all n and a>0

So, applying the comparison test:

$\sum_{n->0}^{\infty}\frac{1}{n^{a/2}}$

Which converges for p>1, thus, for a>2. Great, I just provided proof that it converges for a>2.

While it's obvious that when a=2 it becomes the harmonic series, I can't use that in the proof, because I'm comparing a smaller series with a bigger one, so I can only proof convergence, how would I go about providing proof of divergence?

Thanks in advance.

 

[LCKurtz]

Try making the denominator a little bigger but still keeping it "like" a harmonic series.

 

[ShizukaSm]

Hey LCKurtz!
Thanks again for all your help, I did it, but I did using the limit comparison test this time, with:
$b_n=\frac{1}{n}$

It worked flawlessly! Out of curiosity, though did you find another solution? I just ask because you mentioned "making it bigger".

 

[LCKurtz]

Hey LCKurtz!
Thanks again for all your help, I did it, but I did using the limit comparison test this time, with:
$b_n=\frac{1}{n}$

It worked flawlessly! Out of curiosity, though did you find another solution? I just ask because you mentioned "making it bigger".

I used the estimate$$
\frac 1 {\sqrt{n^a+1}+\sqrt{n^a}}\ge \frac 1 {\sqrt{n^a+1}+\sqrt{n^a+1}}
=\frac 1 {2\sqrt{n^a+1}}$$and used the limit comparison on that with $\frac 1 n$ for $a=2$.