Tanwir
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Determine the sum or show that the series diverges:

Infinity
E (n+2)^1/2 - 2(n+1)^1/2 + (n)^1/2
n=1

The E is supposed to be a sigma. I'm not too sure how to go about this.

Thanks in advance.
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Farhan.Hanif93
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(Original post by Tanwir)
Determine the sum or show that the series diverges:

Infinity
E (n+2)^1/2 - 2(n+1)^1/2 + (n)^1/2
n=1

The E is supposed to be a sigma. I'm not too sure how to go about this.

Thanks in advance.
Note (n+2)^{1/2} - 2(n+1)^{1/2} + n^{1/2} \equiv ((n+2)^{1/2} - (n+1)^{1/2}) - ((n+1)^{1/2} - n^{1/2}). Then consider the Nth partial sum and investigate the limit as N gets large.
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Tanwir
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I'm a little confused. Will finding the limit give me the sum?
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SimonM
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Yes
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Tanwir
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After plugging in some arbitrary values up to n=k, I see that at n=k I'm left with:

1- (2)^1/2 + (k+2)^1/2 - (k+1)^1/2

If I find the limit as k goes to infinity am I not left with: 1 - (2)^1/2 + infinity - infinity?

But can infinity even cancel like that?
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jassi1
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(Original post by Tanwir)
After plugging in some arbitrary values up to n=k, I see that at n=k I'm left with:

1- (2)^1/2 + (k+2)^1/2 - (k+1)^1/2

If I find the limit as k goes to infinity am I not left with: 1 - (2)^1/2 + infinity - infinity?

But can infinity even cancel like that?
The limit of {sqrt(k+2)-sqrt(k+1)} is zero. Intuitively as k gets extremely large sqrt(k+1) is approximately equal to sqrt(k+2)

You can prove this expression tends to zero by noting:

 \sqrt{n+2}-\sqrt{n+1} = \frac{\sqrt{n+2}^2-\sqrt{n+1}^2}{\sqrt{n+2}+\sqrt{n  +1}}

by difference of two squares, which equals

 \frac{1}{\sqrt{n+2}+\sqrt{n+1}} for integer n

and then you could use the sandwich theorem to show it tends to zero.
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Farhan.Hanif93
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(Original post by Tanwir)
After plugging in some arbitrary values up to n=k, I see that at n=k I'm left with:

1- (2)^1/2 + (k+2)^1/2 - (k+1)^1/2

If I find the limit as k goes to infinity am I not left with: 1 - (2)^1/2 + infinity - infinity?

But can infinity even cancel like that?
Note (k+2)^{1/2} - (k+1)^{1/2} \equiv (k+2)^{1/2} - (k+1)^{1/2}\times \dfrac{(k+2)^{1/2} + (k+1)^{1/2}}{(k+2)^{1/2} + (k+1)^{1/2}}

EDIT: too slow.
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