# Sigma Proof by Induction

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• Aug 31st 2012, 06:20 PM
renolovexoxo
Sigma Proof by Induction
Prove 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1

I've been using induction to prove this.

For the base case I got: 1/1(2)=1/2

Then I go on and get a little more stuck:

Assume 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1 for all n in N. Want to show that 1/1(2)+2/2(3)+...+1/n(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
I simplified it using the assumption

n/(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
Then when I tried to simplify I keep getting something that isn't an equality. Have I done something wrong so far?
• Sep 1st 2012, 01:02 AM
Prove It
Re: Sigma Proof by Induction
Quote:

Originally Posted by renolovexoxo
Prove 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1

I've been using induction to prove this.

For the base case I got: 1/1(2)=1/2

Then I go on and get a little more stuck:

Assume 1/1(2)+2/2(3)+...+1/n(n+1)=n/n+1 for all n in N. Want to show that 1/1(2)+2/2(3)+...+1/n(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
I simplified it using the assumption

n/(n+1)+1/(n+1)(n+2)=(n+1)/(n+1)+1
Then when I tried to simplify I keep getting something that isn't an equality. Have I done something wrong so far?

Notice that \displaystyle \begin{align*} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \end{align*}, so your sum

\displaystyle \begin{align*} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} &= \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 + 1}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n}{n+1} \end{align*}
• Sep 1st 2012, 09:20 AM
renolovexoxo
Re: Sigma Proof by Induction
Quote:

Originally Posted by Prove It
Notice that \displaystyle \begin{align*} \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1} \end{align*}, so your sum

\displaystyle \begin{align*} \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n + 1)} &= \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 + 1}{n + 1} - \frac{1}{n + 1} \\ &= \frac{n}{n+1} \end{align*}

I have to prove by induction though, so I couldn't use this approach, correct?
• Sep 1st 2012, 10:55 AM
Plato
Re: Sigma Proof by Induction
Quote:

Originally Posted by renolovexoxo
I have to prove by induction though, so I couldn't use this approach, correct?

Here is what is needed.
$\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n^2+2n+1} {(n+1)(n+2)}=\frac{n+1}{(n+2)}$
• Sep 1st 2012, 08:33 PM
Prove It
Re: Sigma Proof by Induction
Quote:

Originally Posted by renolovexoxo
I have to prove by induction though, so I couldn't use this approach, correct?

That's not what you said. You said that you were trying to prove the statement and that you used an induction approach. I just showed you an easier approach...
• Sep 2nd 2012, 08:24 AM
renolovexoxo
Re: Sigma Proof by Induction
I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.
• Sep 2nd 2012, 08:56 AM
Plato
Re: Sigma Proof by Induction
Quote:

Originally Posted by renolovexoxo
I'm sorry, then I misstated my intentions. I need to use induction to complete this proof.

Did you follow reply #4?
• Sep 2nd 2012, 05:05 PM
renolovexoxo
Re: Sigma Proof by Induction
I understood it, if that is what you're asking.