# Induction proof

## Homework Statement

This is my first proof by induction so I need some assistance

If n is a positive integer, then $$\sum1/(k(k+1))$$from k=1 to n is equal to n/(n+1)

## Homework Equations

I'm not sure if this is useful for this proof but we are given the proposition:
let n be a positive integer. Then the sum of the first n positive integers is equal to (n(n+1))/2.

## The Attempt at a Solution

(1) Let S be the set of all positive integers k such that 1/2+1/6+...+1/k(k+1)= n/(n+1)
(2) 1 is in S because 1/1(1+1)=1/2 which is in S (not sure if this is the correct approach)
(3) Let n be any element of S and let t=n+1

Am I on the right track? For my next step would I just substitute t=n+1 and solve?
Thanks

Homework Helper
Try decomposing

$$\frac{1}{k(k+1)}$$

into partial fractions, and writing your original sum in terms of those partial fractions.

Thank you statdad I believe this helped but am still curious about my step (2).
Here is what I have changed:

(1)Let S be the set of all positive integers k such that 1/2+(1/2)(1/3)+...+(1/k)(k+1)= n/(n+1)
(2) 1 is in S because (1/1)(1/2)=1/2 which is in S (not sure if this is the correct approach)
(3) Let n be any element of S and let t=n+1
(4) (1/1)(1/2)+(1/2)(1/3)+...+(1/n)(1/n+1)+(1/t)(1/t+1)=(n/(n+1))+(1/t)(1/t+1)
(5) n=t-1
(6) Substituting this into the right hand side leads to: t/(t+1)

I'm pretty sure that's the way to do it I did the algebra for step 6 but did not want to include all the steps. Is (2) correctly justified?

Thanks again

Dick
Homework Helper
Step (2) in your proof is hard to argue with. There's only one term in the sum which is 1/(1*2) And your formula for the sum of the series F(n)=n/(n+1) is also 1/2. I'm not really following what you did for the other steps. You want to prove F(n+1)-F(n)=1/((n+1)*(n+2)), right? Since that's the extra term you added on to the series. I'm pretty unclear on what algebra you actually did. BTW statdad was pointing out a noninductive way to prove it using telescoping series.

Last edited:
No, I do not want to prove F(n+1)-F(n)=1/((n+1)*(n+2)).

Here is my algebra for my last step, maybe it will clarify things:

once we let n=t-1 and plug it into (n/(n+1))+(1/t)(1/t+1) we get the following:
((t-1)/t)+(1/(t(t+1))) finding a common denominator gives ((t+1)(t-1)+1)/(t(t+1))

Multiplying out the numerator gives t2/(t(t+1))

A t in the numerator and denominator cancel giving t/(t+1)
Therefore t is in the form we are looking to prove therefore meaning that t=n+1 exists in S.

Does this make sense now?

Dick