Need help with Mathematical induction question?

**darcangeloel** · October 29th 2012, 05:56 PM

I'm working on a problem and I'm not sure I have this correct on how to do it.
here is the problem
Prove the following statement by mathematical induction:
1/2+1/6+...+1/n(n+1)=n/n+1, for all integers n> or = 1
My first thought is to solve the n/n+1 side with the number 1
So I get 1/2 on both sides by substituting. Then I start to become unsure on where to go from there.
Any help would be awesome.

**MarkFL** · October 29th 2012, 06:08 PM

First show the base case $P_1$ is true:

$\frac{1}{2}=\frac{1}{1(1+1)}=\frac{1}{2}$

True.

State the induction hypothesis $P_n$ :

$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$

Add to this the equation:

$\frac{1}{(n+1)((n+1)+1)}=\frac{1}{(n+1)(n+2)}$

Simplify by incorporating the added term within the summation on the left and combining terms on the right, and you will find you have derived $P_{n+1}$ from $P_n$ , thereby completing the proof by induction.

**Prove It** · October 29th 2012, 06:09 PM

Induction is not necessary.

$\displaystyle \begin{align*} \sum_{k = 1}^n {\frac{1}{k(k + 1)}} &= \sum_{k = 1}^n{\left( \frac{1}{k} - \frac{1}{k + 1} \right)} \\ &= \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} + \frac{1}{n + 1} \right) \\ &= 1 - \frac{1}{n + 1} \\ &= \frac{n}{n + 1} \end{align*}$

**darcangeloel** · October 29th 2012, 06:12 PM

Mark, awesome answer, but a quick question I thought you had to go prove it in terms of K+1. Although I've been wrong before, thanks a bunch.

**MarkFL** · October 29th 2012, 06:20 PM

If you carry out the operations I suggested you will get:

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{n+1}{n+2}= \frac{n+1}{(n+1)+1}$

This is $P_{n+1}$ , which was derived from $P_n$ .

**darcangeloel** · October 29th 2012, 06:50 PM

Wait, so how did we get 1/(n+1)((n+1)+1) = n+1/n+2 sorry for drawing this question out just trying to understand the different parts.

**MarkFL** · October 29th 2012, 10:20 PM

State the induction hypothesis $P_n$ :

$\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$

Add to this the equation:

$\frac{1}{(n+1)((n+1)+1)}=\frac{1}{(n+1)(n+2)}$

$\sum_{k=1}^n\frac{1}{k(k+1)}+\frac{1}{(n+1)((n+1)+ 1)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}$

On the left we may incorporate the added term within the summation:

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{n}{n+1}+ \frac{1}{(n+1)(n+2)}$

On the right combine terms:

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(n+\frac{1}{n+2} \right)$

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(\frac{n(n+2)+1}{n+2} \right)$

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{1}{n+1} \left(\frac{n^2+2n+1}{n+2} \right)$

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{(n+1)^2}{(n +1)(n+2)}$

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{n+1}{n+2}$

$\sum_{k=1}^{n+1}\frac{1}{k(k+1)}=\frac{n+1}{(n+1)+ 1}$

Now we have $P_{n+1}$ , completing the proof by induction.

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