Need help with Mathematical induction question?

• Oct 29th 2012, 05:56 PM
darcangeloel
Need help with Mathematical induction question?
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.
• Oct 29th 2012, 06:08 PM
MarkFL
Re: Need help with Mathematical induction question?
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.
• Oct 29th 2012, 06:09 PM
Prove It
Re: Need help with Mathematical induction question?
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*}
• Oct 29th 2012, 06:12 PM
darcangeloel
Re: Need help with Mathematical induction question?
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.
• Oct 29th 2012, 06:20 PM
MarkFL
Re: Need help with Mathematical induction question?
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 $\displaystyle P_n$.
• Oct 29th 2012, 06:50 PM
darcangeloel
Re: Need help with Mathematical induction question?
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.
• Oct 29th 2012, 10:20 PM
MarkFL
Re: Need help with Mathematical induction question?
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.