How much is ∑(1/(√j + √(j + 1)) (j=0..n)?

The response is√(n+1).

To simplify, you have ∑a_j

Multiply numerator and denominator in a_j by √(j+1) - √j. You will see that a_j is actually √(j+1) - √j. When you do the sum, √(j+1) from a_j is canceled by -√(j+1) of a_(j+1).

Rationalize the denominator:

1 / [√(j+1) + √j] = [√(j+1) - √j] / [(j+1) - j]

1 / [√(j+1) + √j] = √(j+1) - √j

In a sum, the positive part of each term is canceled by the negative part of the next term. With the sum from j=0 to n:

∑ 1 / [√(j+1) + √j] = [√1 - √0] + [√2 - √1] + ... + [√(n+1) - √n]

= √(n+1) - √0

∑ 1 / [√(j+1) + √j] = √(n+1)

Using a convergence test letting n go off to infinity, you will find that your equation doesn't converge to a particular number. We would say that your series is 'divergent'

However if you used a partial sum formula , you will find that your above equation settles on the value √(n+1)

Hope this helps

What the fvck is that? O.O

j=1028 just kidding... i dont know...