What is the sum of the first n terms of 1/(n(n+1))?

Question

I already know the solution, (1 - 1/(n+1)) so if you could explain how this is found it would be appreciated. I see this as having some harmonic series in it.

Answer 1

Use simple fraction and you will get a telescopic serie.

1/n(n+1) = 1/n - 1/(n+1)

So: you have this.

(1/1 -1 /2) + (1/2 - 1/3) + ... + (1/n - 1/n+1)

Just use the fact that 2 opposite terms sum is 0 =)

Ana

Answer 2

Use partial fractions

1/(n(n+1)) = 1/n-1/(n+1)

Let's say you want k terms of 1/(n(n+1)), starting from n = 1 to n = k. then, we have:

1/1-1/2+1/2-1/3+1/3-1/4... +1/(k-1)-1/k+1/k-1/(k+1)
=1-1/(k+1)

As Ana said, this series telescopes which means that the 1/(n+1) part of one term cancels out with the -1/n part of the next term as you go from n = x to n = x+1

Answer 3

Since 1/(n(n+1)) equals 1/n - 1/(n+1),
adding the first n terms alternately adds and subtracts
the same terms.

So we get

sum = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

= 1 + 0 + 0 + 0 + ... + 0 - 1/(n+1) = 1 - 1/(n+1).

As stated.

Answer 4

Nope, no harmonic series involved here. It's a simple, monotonically increasing function on n which converges to 1 as n -> ∞.


Doug

Answer 5

the first n is n=1? 'cause it might be n=0