Finite sum where every term is a square root

Question

How do I compute

$$\sum_{r=1}^{2000}{\sqrt{1 + \frac1{r^2} + \frac1{(r+1)^2}}}$$ I have tried writing inside term as a perfect square but leading nowhere.

Answer 1

Hint: Use that your radicand is given by $${\frac { \left( {r}^{2}+r+1 \right) ^{2}}{{r}^{2} \left( r+1 \right) ^ {2}}} $$

Answer 2

Hint $${\sqrt{1 + \frac1{r^2} + \frac1{(r+1)^2}}}=\sqrt{\frac{(1 + r + r^2)^2}{r^2 (1 + r)^2}}=\frac{(1 + r + r^2)}{r(r+1)}=1+ \frac1r -\frac1{r + 1} $$

Answer 3

We have $$\displaystyle \sqrt{1 + \frac{1}{n^2}+\frac{1}{(n+1)^2}} = \sqrt{\frac{(n^2+n+1)^2}{n^2(n+1)^2}} = \frac{n^2+n+1}{n(n+1)} = 1 + \frac 1n - \frac{1}{n+1}$$

So, $$\sum_{n=1}^{2000} \left( 1 + \frac 1n - \frac{1}{n+1}\right) = \underbrace{\sum_{n=1}^{2000}1}_{=2000} + \underbrace{\sum_{n=1}^{2000} \left( \frac 1n- \frac{1}{n+1}\right)}_{=?}$$

Answer 4

We need to prove that

$$1+\dfrac{1}{r^2}+\dfrac{1}{(r+1)^2}=\left(1+\dfrac{1}{r(r+1)}\right)^2$$

$1+\dfrac{1}{r^2}+\dfrac{1}{(r+1)^2}=1+\dfrac{r^2+(r+1)^2}{r^2(r+1)^2}=1+\dfrac{2r^2+2r+1}{r^2(r+1)^2}$

$=1+\dfrac{2r(r+1)}{(r(r+1))^2}+\dfrac{1}{r^2(r+1)^2}=1+2\times \dfrac{1}{r(r+1)}+ \left(\dfrac{1}{r(r+1)}\right)^2=\left(1+\dfrac{1}{r(r+1)}\right)^2$

Answer 5

Here:$${\sqrt{1 + \frac1{r^2} + \frac1{(r+1)^2}}}=\frac{1 + r + r^2}{r^2+r}=\frac{1}{r}-\frac{1}{r+1}+1=\color{red}{\frac{1}{r(r+1)}}+1$$ So $$\sum_{r=1}^{2000}{\sqrt{1 + \frac1{r^2} + \frac1{(r+1)^2}}}=\sum_{r=1}^{2000}{\color{red}{\frac{1}{r(r+1)}}+2000}=\frac{2000}{2001}+2000\approx2001$$ The trick to get $\frac{2000}{2001}$ is by telescoping sum. Writing some terms:$$\sum_{r=1}^{2000}{\frac{1}{r}-\frac{1}{r+1}}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\dots-\frac{1}{2000}+\frac{1}{2000}-\frac{1}{2001}=1-\frac{1}{2001}$$

Note: The red guy is very close to $1$ when summing to infinity. This line ensures you did not calculate incorrectly.