solve by induction

Question

$$\sum_{r=2}^n{1\over r^2-1}=\frac34-{2n+1\over 2n(n+1)}$$

after I got to $n=k+1$ and tried to get both sides equal I got stuck,

prove: $n=k+1$ ;

$${1\over k^2 -1} + {1\over (k+1)^2 -1}=\frac34 - {2(k+1) +1\over 2(k+1)(k+2)}$$

any help would be appreciated.

The sum is telescopic: $$\frac1{r^2-1}=\frac1{(r-1)(r+1)}=\frac12\bigg(\frac1{r-1}-\frac1{r+1}\bigg)$$ — Lucian, Feb 02 '14 at 16:18
Using the telescopic series suggested by @Lucian, all terms cancel out except $\frac{1}{2}, -\frac{1}{n}, -\frac{1}{n+1}$. And you will get the same result. — Sandeep Thilakan, Feb 02 '14 at 16:27
Check this for $n=2$ and you will find out that it is not correct: $\frac{1}{3}\neq\frac{3}{4}-\frac{2\times2-1}{2\times2\times\left(2+1\right)}$ — drhab, Feb 02 '14 at 16:50
The base case will be true if $\frac{3}{4}$ is replaced by $\frac{7}{12}$. — drhab, Feb 02 '14 at 16:55

score 1 · Accepted Answer · answered Feb 02 '14 at 16:09

1

Let $\displaystyle\sum_{r=2}^n{1\over r^2-1}=\frac34-{2n-1\over 2n(n+1)}$ holds true for $n=m$

Let $\implies\displaystyle\sum_{r=2}^m{1\over r^2-1}=\frac34-{2m-1\over 2m(m+1)}$

For $n=m+1,$ $\displaystyle\implies\sum_{r=2}^{m+1}{1\over r^2-1}=\sum_{r=2}^m{1\over r^2-1}+\frac1{(m+1)^2-1}=\frac34-{2m-1\over 2m(m+1)}+\frac1{m(m+2)}$

Now, $\displaystyle\frac1{m(m+2)}-\frac{2m-1}{2m(m+1)}=\frac{2(m+1)-(m+2)(2m-1)}{2m(m+2)(m+1)}=\frac{-m(2m+1)}{2m(m+2)(m+1)}$ $\displaystyle=-\frac{2m+1}{2(m+2)(m+1)}$

Now establish the base case i.e., for $n=1$

answered Feb 02 '14 at 16:09

lab bhattacharjee

274,582

i did all that already :/ still couldn't get it, algebra is lagging atm – luke Feb 02 '14 at 16:15
@luke, have you got your answer? – lab bhattacharjee Feb 02 '14 at 16:16
your numinator in ur answe is not like mine, i got $2(m+1)+1$ – luke Feb 02 '14 at 16:27
@luke, have you identified any mistake with the calculation of mine or yours – lab bhattacharjee Feb 02 '14 at 16:31
you seem to have taken the 2's away instead of adding them maybe? – luke Feb 02 '14 at 16:33
i've gotten it, thanks. you just mistaked the 2's. – luke Feb 02 '14 at 16:37
Base case is not $n=1$ but $n=2$ and if I am not mistaken then the base case fails to be true. – drhab Feb 02 '14 at 16:38
i've worked the base case to be true -> $\frac13$ – luke Feb 02 '14 at 16:44

score 0 · Answer 2 · answered Feb 02 '14 at 16:21

$$\sum_{r=2}^{n+1}\frac{1}{r^{2}-1}=\frac{3}{4}-\frac{2n-1}{2n\left(n+1\right)}+\frac{1}{\left(n+1\right)^{2}-1}=\frac{3}{4}-\left[\frac{2n-1}{2n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\right]=$$$$\frac{3}{4}-\frac{2n+1}{2\left(n+1\right)\left(n+2\right)}$$

solve by induction

2 Answers2