How do you prove 1/(1*2)+1/(2*3)+...+1/(n(n+1)) = 1-1/(n+1) by induction?

3 Answers
Dec 17, 2017

See explanation...

Explanation:

Let P(n) be the proposition:

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

Base case

Note that:

1/(1*2) = 1/2 = 1-1/2 = 1-1/(1+1)

So P(1) is true.

Induction step

Suppose P(n) is true.

Then:

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

=1-1/(n+1)+1/((n+1)(n+2))

=1-((n+2)/((n+1)(n+2))-1/((n+1)(n+2)))

=1-((n+2)-1)/((n+1)(n+2))

=1-(n+1)/((n+1)(n+2))

=1-1/(n+2)

That is P(n) => P(n+1)

Conclusion

We have shown P(1) and P(n)=>P(n+1).

So by induction P(n) holds for all n >= 1

Dec 17, 2017

Show that it is true for n = 1, and show that if it is true for n = k, then it must also be true for n = k + 1, where you may use proof by contradiction.

Explanation:

In mathematical induction, there are two steps:
1. Show that it is true for the first term
2. Show that if it is true for a term k, then it must also be true for a term k + 1 (by first assuming it is true for a term k).

Here is our current sequence:

S_n = 1/(1 * 2) + 1/(2 * 3) + ... + 1/((n - 1)n) + 1/(n(n + 1))
= 1 - (1/(n + 1))

Huh, this feels a bit wrong, there's a similar answered question, but it's not quite the same. Anyway, we should be able to rewrite this sum in terms of sigma:

S_n = sum_(k = 1)^(n) (1/(n(n+1))) = 1 - (1 / (n + 1))

The first step is to show that it is true for the first term. Here we go, n = 1:

S_1 = (1/(1(1+1))) = 1 - (1 / (1 + 1))

S_1 = 1/2 = 1/2

Nice! Turns out this was a different sum than what I thought. For the second step, assume that the sum is true for a constant, say, m:

S_m = sum_(k = 1)^(m) (1/(m(m+1))) = 1 - (1 / (m + 1))

And now, we need to prove that it is true for S_(m + 1):

S_(m + 1) = sum_(k = 1)^(m + 1) (1/((m + 1)(m+2))) = 1 - (1 / (m + 2))

Let's start a little proof by contradiction: assume it isn't, then show that it must be.

S_(m + 1) = sum_(k = 1)^(m + 1) (1/((m + 1)(m+2))) ne 1 - (1 / (m + 2))

The thing is, though, is that S_m and S_(m + 1) share S_m. In other words:

S_(m + 1) = 1/((m + 1)(m+2)) + sum_(k = 1)^(m) (1/(m(m+1)))
ne 1 - (1 / (m + 2))

S_m, as we have assumed, is equal to 1 - (1 / (m + 1)):

S_(m + 1) = 1/((m + 1)(m+2)) + 1 - (1 / (m + 1))
ne 1 - (1 / (m + 2))

At this point we can start ignoring S_(m + 1).

1/((m + 1)(m+2)) + 1 - (1 / (m + 1))
ne 1 - (1 / (m + 2))

We can subtract both sides by 1:

1/((m + 1)(m+2)) - 1 / (m + 1) ne -1 / (m + 2)

Interesting. Let's somehow make all of these fractions share the same denominator:

1/((m + 1)(m+2)) - (m + 2) / ((m + 1)(m + 2))
ne -(m + 1) / ((m + 1)(m + 2))

Combining those two fractions on the left hand side:

(1 - (m + 2))/((m + 1)(m+2)) ne -(m + 1) / ((m + 1)(m + 2))

Subtracting the fraction on the right hand side, by itself:

(1 - (m + 2))/((m + 1)(m+2)) + (m + 1) / ((m + 1)(m + 2)) ne 0

Let's simplify:

(1 - (m + 2) + (m + 1))/((m + 1)(m+2)) ne 0

(1 - m - 2 + m + 1)/((m + 1)(m+2)) ne 0

0/((m + 1)(m+2)) ne 0

0 ne 0

We have a contradiction! Therefore, it is also true that if the sequence is true for a term m, then it must also be true for a term m + 1, and our proof by mathematical induction (and contradiction) is complete.

Dec 17, 2017

Proof via method of differences

Explanation:

I wanted to provide another approach to showing this is true:

We know 1/(1*2) + 1/(2*3) + ... + 1/(n(n+1))

= sum_(r = 1) ^n 1/(r(r+1) )

Using partial fractions:

=> 1/( r(r+1) ) = A/r + B/(r+1)

1 = A(r+1) + Br

Letting r = 0 => A = 1

Letting r = -1 => B = -1

=> = sum_(r=1) ^n 1/r - 1/(r+1)

r =1 : 1/1 cancel(- 1/2)

r = 2: cancel( 1/2) cancel(- 1/3)
.
.
.
r = n-1: cancel(1/(n-1)) cancel(- 1/n)
r = n: cancel(1/n) - 1/(n+1)

Hence using what is left:

1/1 - 1/(n+1)

=> 1 - 1/(n+1)