Sum Of Series Of Square Roots A Step-by-Step Solution

In the realm of mathematics, we often encounter intriguing problems that require a blend of algebraic manipulation and insightful observation. One such problem involves finding the sum of a series of square roots, each term of which has a specific structure. This article delves into the solution of the following problem:

Find the sum of $\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}}, \sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}}, \dots, \sqrt{1 + \frac{1}{20^2} + \frac{1}{21^2}}$

This problem, at first glance, might seem daunting due to the presence of square roots and fractions. However, with a clever algebraic trick and a keen eye for patterns, we can unravel its solution. We will explore the underlying mathematical principles, provide a step-by-step solution, and discuss the broader implications of this problem in the context of mathematical problem-solving.

Dissecting the General Term

The cornerstone of solving this problem lies in understanding the structure of the general term within the series. Let's consider the n-th term of the series, which can be expressed as:

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

Our goal is to simplify this expression, ideally to a form that allows us to easily compute the sum of the series. The key here is to recognize that the expression inside the square root can be manipulated into a perfect square. This is a common technique in mathematical problem-solving, where we try to rewrite complex expressions in simpler, more manageable forms.

To achieve this, we first find a common denominator for the fractions inside the square root:

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

Next, we expand the terms in the numerator:

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

Combining like terms in the numerator, we get:

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

Now, we can rewrite the 1 outside the fraction as $\frac{n^2(n+1)^2}{n^2(n+1)^2}$ and add it to the fraction:

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

Expanding $n^2(n+1)^2$ gives $n^2(n^2 + 2n + 1) = n^4 + 2n^3 + n^2$, so the expression becomes:

$$\sqrt{\frac{n^4 + 2n^3 + n^2 + 2n^2 + 2n + 1}{n^2(n+1)^2}}$$

Combining like terms in the numerator, we have:

$$\sqrt{\frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2}}$$

At this point, the numerator might look intimidating, but with a bit of insight, we can recognize it as a perfect square. Specifically, it is the square of $n^2 + n + 1$. To see this, let's expand $(n^2 + n + 1)^2$:

$$(n^2 + n + 1)^2 = (n^2 + n + 1)(n^2 + n + 1) = n^4 + n^3 + n^2 + n^3 + n^2 + n + n^2 + n + 1 = n^4 + 2n^3 + 3n^2 + 2n + 1$$

This matches the numerator of our expression! Therefore, we can rewrite the general term as:

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

Now, we can take the square root of both the numerator and the denominator:

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

This simplified expression is a significant step forward. We can further break it down by dividing each term in the numerator by $n(n+1)$:

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

Combining the first two terms, we get:

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

Finally, we can decompose the fraction $\frac{1}{n(n+1)}$ using partial fractions. We seek constants A and B such that:

$$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$

Multiplying both sides by $n(n+1)$, we get:

$$1 = A(n+1) + Bn$$

To find A, we can set n = 0: 1 = A(0+1) + B(0) => A = 1

To find B, we can set n = -1: 1 = A(-1+1) + B(-1) => B = -1

Thus,

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

Substituting this back into our expression for the general term, we obtain:

$$1 + \frac{1}{n} - \frac{1}{n+1}$$

This is the simplified form of the general term, and it is crucial for evaluating the sum of the series. The expression 1 + 1/n - 1/(n+1) is the simplified form of the general term. This simplification is key to solving the problem because it reveals a telescoping pattern when summing the series. Now, let's delve into how this telescoping pattern works and how it enables us to find the sum efficiently.

The Telescoping Sum Technique

The expression we derived for the general term, $1 + \frac{1}{n} - \frac{1}{n+1}$, is in a form that lends itself perfectly to the telescoping sum technique. This technique is a powerful tool for evaluating sums where intermediate terms cancel out, leaving only a few terms at the beginning and end of the series. The name