Question

Need some help with mathematical induction please:
We have Sigma r=2 to n for 2/(r(r^2-1))=1/2 - 1/(n(n+1)) for n = 2,3,4...
I know what I need to deduce (I think) but am struggling to make the equations match. I have:
1/2 - 1/((k+1)(k+2)) should equal 1/2 - 1/(k(k+1)) + 2/(k+1)((k+1)^2-1). Getting stuck reducing it down.

Answer 1

Let me help you show the problem
\(\Large \sum_{r=2}^{n}\dfrac{2}{r(r^2-1)}=\dfrac{1}{2}-\dfrac{1}{n(n+1)}\)

Answer 2

but I really don't understand what we are supposed to do. Prove by induction?

Answer 3

yes, prove the above by induction

Answer 4

for p(2) works ok. We can assume p(k) and then try to deduce p(k+1). It's when I'm trying to go from p(k) to p(k+1) that I'm getting stuck, just algebraicly

Answer 5

Under the \(p(k)\) assumption, you have
\[\sum_{r=2}^k\frac{2}{r(r^2-1)}=\frac{1}{2}-\frac{1}{k(k+1)}\]
And you must establish \(p(k+1)\), i.e.
\[\sum_{r=2}^{k+1}\frac{2}{r(r^2-1)}=\frac{1}{2}-\frac{1}{(k+1)(k+2)}\]
Consider the first \(k-1\) terms of the \(p(k+1)\) summation (\(k-1\) terms because that's how many terms you have from 2 to \(k\)):
\[\begin{align*}\sum_{r=2}^{k+1}\frac{2}{r(r^2-1)}&=\sum_{r=2}^{k}\frac{2}{r(r^2-1)}+\frac{2}{(k+1)((k+1)^2-1)}\\
&=\left[\frac{1}{2}-\frac{1}{k(k+1)}\right]+\frac{2}{(k+1)(k^2+2k)}\\
&=\frac{1}{2}-\left[\frac{1}{k(k+1)}-\frac{2}{(k+1)(k^2+2k)}\right]\\
&=\frac{1}{2}-\frac{1}{k(k+1)}\left[1-\frac{2}{k+2}\right]\\
&=\frac{1}{2}-\frac{1}{k(k+1)}\left[\frac{k+2}{k+2}-\frac{2}{k+2}\right]\\
&=\frac{1}{2}-\frac{1}{k(k+1)}\left[\frac{k}{k+2}\right]\\
&=\frac{1}{2}-\frac{1}{(k+1)(k+2)}
\end{align*}\]
And we're done.

Answer 6

Thank you! I was with it all the way up to the fourth line from the bottom. Couldn't work out to take as a fact\[\frac{ 1 }{ k(k+1) }\] as a factor. Then a nice trick I always overlook to make \[1 = \frac{ k+2 }{ k+2 }\].

Thanks for the help! :)

Answer 7

yw!