# Determine the sum of the series

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#1
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.

0
6 years ago
#2
(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.

Note . Then consider the Nth partial sum and investigate the limit as N gets large.
3
#3
I'm a little confused. Will finding the limit give me the sum?
0
6 years ago
#4
Yes
0
#5
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?
0
6 years ago
#6
(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:

by difference of two squares, which equals

for integer n

and then you could use the sandwich theorem to show it tends to zero.
1
6 years ago
#7
(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

EDIT: too slow.
0
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