MHB Integer S most close to A and less than A

[Albert1]

$A=\sqrt{1^2+\dfrac{1}{1^2+2^2}}+\sqrt{1^2+\dfrac{1}{2^2+3^2}}+\sqrt{1^2+\dfrac{1}{3^2+4^2}}+---+\sqrt{1^2+\dfrac{1}{2011^2+2012^2}}$
please find an integer S most close to A and less than A

 

[Albert1]

$A=\sqrt{1^2+\dfrac{1}{1^2+2^2}}+\sqrt{1^2+\dfrac{1}{2^2+3^2}}+\sqrt{1^2+\dfrac{1}{3^2+4^2}}+---+\sqrt{1^2+\dfrac{1}{2011^2+2012^2}}$
please find an integer S most close to A and less than A

sorry :a typo
$A=\sqrt{1^2+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+-------+\sqrt{1^2+\dfrac{1}{2011^2}+\dfrac{1}{2012^2}}$

 

[Greg]

Spoiler

$$A=\sum_{n=1}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sum_{n=1}^{2011}\sqrt{\frac{[n(n+1)]^2+(n+1)^2+n^2}{[n(n+1)]^2}}$$

$$=\sum_{n=1}^{2011}\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\sqrt{\frac{(n^2+n+1)^2}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\frac{n^2+n+1}{n(n+1)}$$

$$=\sum_{n=1}^{2011}\left(1+\frac{1}{n(n+1)}\right)=2011+\sum_{n=1}^{2011}\left(\frac1n-\frac{1}{n+1}\right)\Rightarrow S=2011$$

 

[Albert1]

Spoiler

$$A=\sum_{n=1}^{2011}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sum_{n=1}^{2011}\sqrt{\frac{[n(n+1)]^2+(n+1)^2+n^2}{[n(n+1)]^2}}$$

$$=\sum_{n=1}^{2011}\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\sqrt{\frac{(n^2+n+1)^2}{[n(n+1)]^2}}=\sum_{n=1}^{2011}\frac{n^2+n+1}{n(n+1)}$$

$$=\sum_{n=1}^{2011}\left(1+\frac{1}{n(n+1)}\right)=2011+\sum_{n=1}^{2011}\left(\frac1n-\frac{1}{n+1}\right)\Rightarrow S=2011$$

great ! your answer is correct