Find the sum of first twenty terms of the following series 1+1/(1^0.5+2^0.5)+1/(2^0.5+3^0.5)+1/(3^0.5+4^0.5)..

Question

The series is 1/(1^0.5+2^0.5)+1/(2^0.5+3^0.5)+1/(3^0.5+4^0.5)+......+1/{n^0.5+(n+1)^0.5}+...
What is the sum of first twenty terms of this series.
Note: root over n is written as n^0.5

The full series is not shown. Put your mouse pointer over the dots at the end of the series, the remaining terms of the series will be shown.

Answer 1

Sum of the first twenty terms = sqrt(20)

Each term may be represented by:
1 / [sqrt(m) + sqrt(n)], where n > m.

Multiply this by
[sqrt(n) - sqrt(m)] / [sqrt(n) - sqrt(m)]
and you get:
[sqrt(n) - sqrt(m)] / (n - m)
But n - m = 1 for each term,
so each term = sqrt(n) - sqrt(m).

The series then becomes:
1 + [sqrt(2) - sqrt(1)] + [sqrt(3) - sqrt(2)] + [sqrt(4) - sqrt(3)] + ...

Sum of 1st term = 1
Sum of first 2 terms = sqrt(2)
Sum of first 3 terms = sqrt(3)
etc.
Sum of first 20 terms = sqrt(20)

In general, sum of first r terms = sqrt(r)