Need help with Mathematical induction question?

Oct 2012
3
0
Richmond VA
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.
 
Dec 2011
2,314
916
St. Augustine, FL.
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 \(\displaystyle P_n\):

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

Add to this the equation:

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

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
Aug 2008
12,897
5,001
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 2012
3
0
Richmond VA
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.
 
Dec 2011
2,314
916
St. Augustine, FL.
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 2012
3
0
Richmond VA
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.
 
Dec 2011
2,314
916
St. Augustine, FL.
State the induction hypothesis \(\displaystyle P_n\):

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

Add to this the equation:

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

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

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.