How to show that this is divergent?

[stukbv]

Homework Statement

Basically i have got the infinite sum of
(n+2)^1/2-n^1/2

and i think it is divergent ( I hope) but i have no idea how to show it, the ratio test is not helpful and i cannot find anything to compare it with.

Thanks in advance.

 

[rock.freak667]

It looks like a telescoping sum, so you can try to sub in some numbers and see what cancels out and what does not until a final value N and then take the limit as N→∞.

 

[stukbv]

Ok thanks but.. What is a telescoping sum?

 

[rock.freak667]

Ok thanks but.. What is a telescoping sum?

http://en.wikipedia.org/wiki/Telescoping_series"

It is essential using the fact that if you need to sum something, terms can cancel out to simplify the sum. The wiki article explains it better than I can with its example.

 

[stukbv]

Thanks! :)

 

[flyingpig]

Why can't you just use the test for divergence? Doesn't it go to infinity immediately?

 

[Dick]

Why can't you just use the test for divergence? Doesn't it go to infinity immediately?

No. sqrt(n+2)-sqrt(n) is positive and approaches zero. So the series is sum of positive terms that go to zero. Is the sum of (n+1/2^n)-n divergent? stukbv could also have multiplied by the conjugate in the numerator and denominator and used a comparison with a p-series to show it's divergent. It is a delicate point. Regrouping a divergent series can make it converge. 1-1+1-1+1-1+... diverges. (1-1)+(1-1)+(1-1)+... converges. You can't just split stuff and move it around like in a finite sum.

 

[shakgoku]

Multiply and divide by
(n+2)^{\frac{1}{2}}+n^{\frac{1}{2}}

After simplification you will get,

\frac{2}{n^{\frac{1}{2}}[1+(1+\frac{2}{n})]}

Now use the comparision test, by comparing with \sum \frac{1}{n^{\frac{1}{2}}} which diverges by P-test.

 

[Maxter]

If you want to use telescoping series, you'll find that the nth partial sum is -1+(n+1)^(1/2)+(n)^(1/2) which diverges to infty.