# How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$

How to find the sum of the following :

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$

Please suggest as getting no clue on this... thanks..

• What do you mean by find exactly? Commented Aug 7, 2013 at 16:48
• $\frac{3999999}{2000}$,by induction Commented Aug 7, 2013 at 16:48

Hint:

Let us write it as follows:

$$\sum_{n=1}^{1999} \sqrt{1+ \frac{1}{n^{2}} + \frac{1}{(n+1)^{2}}}$$

We can rewrite the radical as:

$$\sqrt{\frac{n^{2}(n+1)^{2} + n^{2} + (n+1)^{2}}{n^{2}(n+1)^{2}}}$$

Which simplifies to:

$$\sqrt{\frac{(n^{2}+n+1)^{2}}{n^{2}(n+1)^{2}}} = \frac{n^{2}+n+1}{n^{2}+n} = 1 + \frac{1}{n^{2}+n}$$

So now our sum is:

$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n}$$

Extracting the constant term gives

$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n} = 1999 + \sum_{n=1}^{1999}\frac{1}{n(n+1)}$$

By partial fractions decomposition, we see:

$$\frac{1}{n(n+1)} = \frac{1}{n} -\frac{1}{n+1}$$

The series therefore telescopes, making the sum easy to compute.

• AWertheim, could you explain to me how you factored $n^2 \cdot (n+1)^2 + n^2 + (n+1)^2$ ?
– user158108
Commented Jun 19, 2014 at 23:08
• Hello Pedro. The way I was able to see it was by observing that $n^{2}(n+1)^{2} = n^{4} + 2n^{3} + n^{2}$, so $n^{2}(n+1)^{2} + n^{2} + (n+1)^{2} = n^{4} +2n^{3}+2n^{2} + (n+1)^{2} = (n^{2})^{2} + 2(n^{2})(n+1) + (n+1)^{2}$. From here, I recognized this as the expansion of $(a+b)^{2} = a^{2}+2ab+b^{2}$ where $a = n^{2}$ and $b = (n+1)$, hence the factorization obtained. Hope this helps! Commented Jun 19, 2014 at 23:29
• @user158108: $$n^2 \cdot (n+1)^2 + n^2 + (n+1)^2=n^4+2 n^3+3 n^2+2 n+1$$ But polynomial equations of type $$x^4+ax^3+bx^2+ax+1=0$$ can be transformed to $$(x^4+1)+ax(x^2+1)+bx=0$$ and further to $$x^2((x+\frac{1}{x})^2-2)+a(x+\frac{1}{x})+c)=0$$ Now the substitution $$y=x+\frac{1}{x}$$ can be used. Commented Nov 24, 2014 at 1:26

In addition to AWertheim's answer, in case you haven't studied telescopic series yet, this is how you can calculate $\displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n}$:

First you see that you can write: $\displaystyle \frac{1}{n^{2}+n} = \frac{1}{n} - \frac{1}{n+1}$

This technique is called partial fraction decomposition if you don't know it already.

Now see what happens:

$\displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n} = \displaystyle \sum_{n=1}^{1999}(\frac{1}{n} - \frac{1}{n+1}) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + \cdot \cdot \cdot +(1/1999 - 1/2000) = 1 + (-1/2+1/2) + (-1/3 +1/3) +\cdot \cdot \cdot + (-1/1999+1/1999)-1/2000= 1 - 1/2000 = \frac{1999}{2000}$ So, each term cancels the previous term, except the first and last terms. This kind of series is called telescopic series.

Also note that: $\displaystyle \sum_{n=1}^{1999} 1 = 1999$

$\displaystyle \sum_{n=1}^{1999} (1 + \frac{1}{n^{2}+n}) = \displaystyle \sum_{n=1}^{1999} 1 + \displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n} = 1999+ \frac{1999}{2000} = \frac{3999999}{2000}$