# Need help with Mathematical induction question?

#### darcangeloel

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

First show the base case $$\displaystyle P_1$$ is true:

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

True.

State the induction hypothesis $$\displaystyle P_n$$:

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

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

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 $$\displaystyle P_{n+1}$$ from $$\displaystyle P_n$$, thereby completing the proof by induction.

#### Prove It

MHF Helper
Induction is not necessary.

\displaystyle \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

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

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

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

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

#### darcangeloel

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

State the induction hypothesis $$\displaystyle P_n$$:

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

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

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

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

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

On the right combine terms:

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

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

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

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

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

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

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