Understanding the Series Sum Problem

The problem asks for the sum of a series of terms, where each term is in the form of a square root expression. The series is given by:

\[ S = \sqrt {1 + \frac{1}{{{1^{2}}}} + \frac{1}{{{2^{2}}}}} + \sqrt {1 + \frac{1}{{{2^{2}}}} + \frac{1}{{{3^{2}}}}} + \ldots \ldots .. + \sqrt {1 + \frac{1}{{{{2007}^{2}}}} + \frac{1}{{{{2008}^{2}}}}} \]

We can see that the general term of the series can be written as:

\[ T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}} \]

where \(n\) ranges from 1 to 2007.

Simplifying the General Term

Let's try to simplify the expression inside the square root for the general term \(T_n\). We look for a way to express \(1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}\) as a perfect square. Consider the expression \(1 + \frac{1}{n} - \frac{1}{n+1}\). Let's square this expression:

\[ \left(1 + \frac{1}{n} - \frac{1}{n+1}\right)^2 = \left(1 + \left(\frac{n+1 - n}{n(n+1)}\right)\right)^2 = \left(1 + \frac{1}{n(n+1)}\right)^2 \]

Expanding this, we get:

\[ \left(1 + \frac{1}{n(n+1)}\right)^2 = 1^2 + 2 \times 1 \times \frac{1}{n(n+1)} + \left(\frac{1}{n(n+1)}\right)^2 = 1 + \frac{2}{n(n+1)} + \frac{1}{n^2(n+1)^2} \]

This does not match the expression inside the square root.

Let's try squaring \(1 + \frac{1}{n} - \frac{1}{n+1}\) using the \( (a+b+c)^2 \) formula, where \( a=1 \), \( b=\frac{1}{n} \), \( c=-\frac{1}{n+1} \):

\[ (1 + \frac{1}{n} - \frac{1}{n+1})^2 = 1^2 + \left(\frac{1}{n}\right)^2 + \left(-\frac{1}{n+1}\right)^2 + 2(1)\left(\frac{1}{n}\right) + 2(1)\left(-\frac{1}{n+1}\right) + 2\left(\frac{1}{n}\right)\left(-\frac{1}{n+1}\right) \]

\[ = 1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} + \frac{2}{n} - \frac{2}{n+1} - \frac{2}{n(n+1)} \]

Now, let's combine the terms with \(n\) and \(n+1\) in the denominator:

\[ \frac{2}{n} - \frac{2}{n+1} - \frac{2}{n(n+1)} = \frac{2(n+1) - 2n}{n(n+1)} - \frac{2}{n(n+1)} = \frac{2n+2 - 2n}{n(n+1)} - \frac{2}{n(n+1)} = \frac{2}{n(n+1)} - \frac{2}{n(n+1)} = 0 \]

So, we find that:

\[ \left(1 + \frac{1}{n} - \frac{1}{n+1}\right)^2 = 1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} \]

Since \(n \ge 1\), the expression \(1 + \frac{1}{n} - \frac{1}{n+1} = 1 + \frac{n+1-n}{n(n+1)} = 1 + \frac{1}{n(n+1)}\) is always positive. Therefore, the square root is:

\[ T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}} = 1 + \frac{1}{n} - \frac{1}{n+1} \]

Evaluating the Sum of the Series

Now we need to find the sum of the series from \(n=1\) to \(n=2007\):

\[ S = \sum_{n=1}^{2007} \left(1 + \frac{1}{n} - \frac{1}{n+1}\right) \]

We can split this sum into two parts:

\[ S = \sum_{n=1}^{2007} 1 + \sum_{n=1}^{2007} \left(\frac{1}{n} - \frac{1}{n+1}\right) \]

The first part is simply adding 1 for 2007 times:

\[ \sum_{n=1}^{2007} 1 = 2007 \times 1 = 2007 \]

The second part is a telescoping series:

\[ \sum_{n=1}^{2007} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{2007} - \frac{1}{2008}\right) \]

In a telescoping series, the intermediate terms cancel out. The sum simplifies to the first term of the first parenthesis and the last term of the last parenthesis:

\[ \sum_{n=1}^{2007} \left(\frac{1}{n} - \frac{1}{n+1}\right) = \frac{1}{1} - \frac{1}{2008} = 1 - \frac{1}{2008} \]

Now, we combine the two parts of the sum:

\[ S = 2007 + \left(1 - \frac{1}{2008}\right) = 2007 + 1 - \frac{1}{2008} = 2008 - \frac{1}{2008} \]

Comparing with Options

The calculated sum is \(2008 - \frac{1}{2008}\). Let's compare this with the given options:

The calculated sum matches Option 1.

Revision Table: Key Concepts

Additional Information: Telescoping Series Explained

A telescoping series is a series whose partial sums have a fixed number of terms after cancellation. This technique is very useful for finding the exact sum of a series when the terms can be expressed as the difference of consecutive terms of a sequence.

For a series \( \sum_{n=1}^{N} a_n \), if \( a_n = f(n) - f(n+1) \), the sum is:

\[ \sum_{n=1}^{N} (f(n) - f(n+1)) = (f(1) - f(2)) + (f(2) - f(3)) + \ldots + (f(N) - f(N+1)) \]

All intermediate terms \(-f(2)\) and \(+f(2)\), \(-f(3)\) and \(+f(3)\), etc., cancel out. The sum reduces to:

\[ f(1) - f(N+1) \]

In our problem, the term was \(1 + \frac{1}{n} - \frac{1}{n+1}\). The summation was \(\sum_{n=1}^{2007} (1 + (\frac{1}{n} - \frac{1}{n+1}))\). The first part \( \sum 1 \) gave \(2007\). The second part \( \sum (\frac{1}{n} - \frac{1}{n+1}) \) is a telescoping series with \( f(n) = \frac{1}{n} \). The sum is \( f(1) - f(2007+1) = f(1) - f(2008) = \frac{1}{1} - \frac{1}{2008} \).

This powerful technique allows us to find the exact sum of series that would otherwise be difficult to evaluate directly.