Telescoping Series: Solving Summations with Alternating Signs

Understanding how to solve summations with alternating signs can be a valuable skill in mathematics. This article explores a problem involving a telescoping series with alternating terms, presenting a detailed step-by-step guide to arrive at the solution.

Introduction to the Problem

Consider the summation of the series:

(S displaystyle sum_{n1}^{30} (-1)^n Big(frac{1}{n} frac{1}{n 1}Big))

This is a complex series with alternating signs and fractions. Our goal is to evaluate this sum step-by-step.

Breaking Down the Problem

Let's start by evaluating the series with a simpler upper bound to understand the pattern:

Step 1: Simplifying with a Lower Upper Bound

First, consider the case when the upper bound is 1:

(displaystyle sum_{n1}^1 (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -Big(1 frac{1}{2}Big) -1 - frac{1}{2})

Next, evaluate the series up to 2:

(displaystyle sum_{n1}^2 (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -Big(1 frac{1}{2}Big) - Big(frac{1}{2} frac{1}{3}Big) -1 - frac{1}{3})

Finally, evaluate the series up to 3:

(displaystyle sum_{n1}^3 (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -Big(1 frac{1}{2}Big) - Big(frac{1}{2} frac{1}{3}Big) - Big(frac{1}{3} frac{1}{4}Big) -1 - frac{1}{4})

From these evaluations, we can conjecture that:

(displaystyle sum_{n1}^m (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -1 - frac{1}{m 1})

Let's prove this conjecture by mathematical induction.

Proof by Mathematical Induction

Base Case: When (m 1), we have:

(displaystyle sum_{n1}^1 (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -Big(1 frac{1}{2}Big) -1 - frac{1}{2})

This matches our conjecture for (m 1).

Inductive Step: Assume the statement holds for (m), i.e.,

(displaystyle sum_{n1}^{m} (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) -1 - frac{1}{m 1})

Now, we need to show it is true for (m 1).

Step 2: Evaluating for (m 1)

Consider:

(displaystyle sum_{n1}^{m 1} (-1)^n Big(frac{1}{n} frac{1}{n 1}Big))

This can be written as:

(displaystyle sum_{n1}^{m} (-1)^n Big(frac{1}{n} frac{1}{n 1}Big) - (-1)^{m 1} Big(frac{1}{m 1} frac{1}{m 2}Big))

Using the inductive hypothesis:

(displaystyle -1 - frac{1}{m 1} - (-1)^{m 1} Big(frac{1}{m 1} frac{1}{m 2}Big))

For (m) even, ((-1)^{m 1} -1) and:

(displaystyle -1 - frac{1}{m 1} Big(frac{1}{m 1} frac{1}{m 2}Big) -1 - frac{1}{m 2})

For (m) odd, ((-1)^{m 1} 1) and:

(displaystyle -1 - frac{1}{m 1} - Big(frac{1}{m 1} frac{1}{m 2}Big) -1 - frac{1}{m 2})

Hence, the statement holds for (m 1) and by induction, the conjecture is true for all positive integers.

Evaluating the Problem with Specific Values

To evaluate the specific problem with (m 30):

(displaystyle S sum_{n1}^{30} (-1)^n Big(frac{1}{n} frac{1}{n 1}Big))

This series is a telescoping series where most terms cancel out.

Step 3: Evaluating the Telescoping Series

Writing out the first few terms:

(S displaystyle -Big(1 frac{1}{2}Big) - Big(frac{1}{2} frac{1}{3}Big) - Big(frac{1}{3} frac{1}{4}Big) - ... - Big(frac{1}{30} frac{1}{31}Big))

Most terms cancel out, leaving:

(S displaystyle -1 frac{1}{31})

Thus:

(S -frac{30}{31})

Conclusion

We have successfully solved the problem of evaluating a series with alternating signs and fractions. By understanding the pattern and applying mathematical induction, we were able to prove the conjecture and reach the final solution.