Evaluate the sum:
(1/(1x2)) + (1/(2x3)) + (1/(3x4)) + (1/(4x5)) . . . + (1/(99x100))
(show work)

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Request for Question Clarification by mathtalk-ga on 03 May 2004 15:10 PDT

Hi, gyrocopter-ga:

The Comments have provided the correct Answer here, with no charge! 
One has to recognize that the sum can be expressed as a "telescoping"
sum (as acidtest4u-ga shows) in which alternating terms cancel out,
leaving only the beginning and ending terms:

1 - 1/100

If their Answer is clear enough, you may want to simply Close (expire)
this Question.

regards, mathtalk-ga

Since I'm an engineer and not a mathematician, I will suggest a simple
solution that lacks grace, but is nonetheless practical.

We have the:
SUM(1 / (N*(N+1))) for N=1 to 99

Since there are only 99 terms, this is easy to solve. Using Matlab with the code:

test = 0;
for n=1:99
   test = test + 1/(n*(n+1));
end
test

We get the result of 0.99.

Yep, there is a more elegant way !! no need to program

I had seen it in my school cursus so long ago, i took some time to
find it back in my rusty brain

just consider that 
1/(n*(n+1)) = (1/n) -(1/(n+1))
the formula now reads
A=1/(1*2) +1/(2*3) + 1/(3*4)....... +1/(n*(n+1))=
[1-(1/2)] +[(1/2)- (1/3)] +[(1/3)-1/4)] ..... +[(1/n)-(1/(n+1)]


Basically, the opposites are neutralising each other, the only remaining terms are

A=1-1/(n+1)

So for n=99
A= 1-1/100= 0,99