Please help me on this ... thanks
The general term appears to be $\displaystyle \displaystyle \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{n^2(n+1)^2}{n^2(n + 1)^2} + \frac{(n + 1)^2}{n^2(n+1)^2} + \frac{n^2}{n^2(n + 1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{n^2(n+1)^2 + (n + 1)^2 + n^2}{n^2(n + 1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{n^2(n^2 + 2n + 1) + n^2 + 2n + 1 + n^2}{n^2(n+1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{n^4 + 2n^3 + n^2 + 2n^2 + 2n + 1}{n^2(n+1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2}}$
$\displaystyle \displaystyle = \sqrt{\frac{(n^2 + n +1)^2}{n^2(n+1)^2}}$
$\displaystyle \displaystyle = \frac{n^2 + n + 1}{n(n + 1)}$
$\displaystyle \displaystyle = \frac{n^2 + n + 1}{n^2 + n}$
$\displaystyle \displaystyle = 1 + \frac{1}{n^2 + n}$
$\displaystyle \displaystyle = 1 + \frac{1}{n(n + 1)}$
$\displaystyle \displaystyle = 1 + \frac{1}{n} - \frac{1}{n + 1}$.
So your series is
$\displaystyle \displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right)$.
This is a telescopic series... See how you go evaluating it...
Oh no it is not. It is so simple.
$\displaystyle \displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right)$=$\displaystyle \displaystyle\sum_{n = 1}^{1999}1+\sum_{n = 1}^{1999}\left(\frac{1}{n} - \frac{1}{n + 1} \right)$=$\displaystyle 1999+1-\frac{1}{2000}$
i would like to apologize to what i had replied on Prove it's answer... i never meant to be so mean... my point was, i didn't understand what he had been solving, i mean the answer itself is too difficult for me to understand , i was thinking that i would be solving too long like adding from 1 to 1999 consecutive numbers because it is summation ... i didn't know how it would go to 1999 + 1 - 1/2000 .... sorry i really cannot get how it would become like that... sorry hope you guys understand... it was just miscommunication.
thanks
Do you know what "telescoping series" means?
Prove It showed that the series can be written
$\displaystyle \displaytype \sum_{n=1}^{1999}\left(1+ \frac{1}{n}- \frac{1}{n+1}\right)$
That is,
$\displaystyle (1+ 1- \frac{1}{2})+ (1+ \frac{1}{2}- \frac{1}{3})+ (1+ \frac{1}{3}+ \frac{1}{4}+ (1+ \frac{1}{4}- \frac{1}{5}) \cdot\cdot\cdot$
Do you see how the fractions cancel? All except in the last term, $\displaystyle -\frac{1}{2000}$. Of course, you will also have "1" added 1999 times.
I'm sure you mean
$\displaystyle \displaystyle \left(1 + 1 - \frac{1}{2}\right) + \left(1 + \frac{1}{2} - \frac{1}{3}\right) + \left(1 + \frac{1}{3} - \frac{1}{4}\right) + \left(1 + \frac{1}{4} - \frac{1}{5}\right) + \dots + \left(1 + \frac{1}{1999} - \frac{1}{2000}\right)$.
But yes, a telescopic series means that all terms except for the first and last end up cancelling.
Many textbooks call them collapsing sums.
Look at $\displaystyle \left[ {1 - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \left[ {\frac{1}{3} - \frac{1}{4}} \right] + \left[ {\frac{1}{4} - \frac{1}{5}} \right] + \left[ {\frac{1}{5} - \frac{1}
{6}} \right]$.
Remove the brackets.
Every term subtracts off, collapses, except the first and last.
So we have $\displaystyle \sum\limits_{k = 1}^5 {\left[ {\frac{1}{k} - \frac{1}{{k + 1}}} \right]} = 1 - \frac{1}{6}$