# 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..

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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.

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

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