Proving the Sum of 1/k - 1/(k 1) n/(n 1) Using Mathematical Induction

This article will guide you through the process of proving the summation of the series 1/k - 1/(k 1) n/(n 1) using mathematical induction. The process involves proving the base case and then the induction step. We will also explore how this series, known as a telescoping series, allows us to simplify the sum.

Overview of the Problem

Given the mathematical induction problem, we need to prove that the sum of the series (sum_{k1}^n left(frac{1}{k} - frac{1}{k 1}right) frac{n}{n 1}). This series is known as a telescoping series, where most terms cancel out, leaving only the first and the last term of the sum.

Base Case

Let's start by establishing the base case. We need to show that the statement holds true for (n1).

**Base Case (n1):**

Consider the sum for (n1):

[sum_{k1}^1 left(frac{1}{k} - frac{1}{k 1}right) frac{1}{1} - frac{1}{2}]

Since there is only one term, the sum simplifies to:

[1 - frac{1}{2} frac{1}{2}]

The right-hand side for (n1) is:

[frac{1}{1 1} frac{1}{2}]

Hence, the base case holds true.

Induction Step

The induction step involves assuming that the statement is true for some arbitrary positive integer (nm), and then proving that it is also true for (nm 1).

**Induction Hypothesis (Assume True for nm):**

Assume that the statement is true for (nm), which means:

[sum_{k1}^m left(frac{1}{k} - frac{1}{k 1}right) frac{m}{m 1}]

**Induction Step (Prove for nm 1):**

Now, we need to show that the statement holds for (nm 1).

Consider the sum for (nm 1):

[sum_{k1}^{m 1} left(frac{1}{k} - frac{1}{k 1}right)]

This can be split as:

[left(sum_{k1}^m left(frac{1}{k} - frac{1}{k 1}right)right) left(frac{1}{m 1} - frac{1}{m 2}right)]

Using the induction hypothesis, the first part of the sum is equal to (frac{m}{m 1}). Therefore, the sum simplifies to:

[frac{m}{m 1} left(frac{1}{m 1} - frac{1}{m 2}right)]

Notice that the (frac{1}{m 1}) terms cancel out, leaving us with:

[frac{m}{m 1} - frac{1}{m 2}]

Combining the fractions, we get:

[frac{m(m 2) - (m 1)}{(m 1)(m 2)} frac{m^2 2m - m - 1}{(m 1)(m 2)} frac{m^2 m - 1}{(m 1)(m 2)}]

This simplifies to:

[frac{m(m 1)}{(m 1)(m 2)} frac{m}{m 2}]

However, we need to show that this is equal to (frac{m 1}{m 2}), which is:

[frac{m 1}{m 2}]

Therefore, the sum for (nm 1) is:

[sum_{k1}^{m 1} left(frac{1}{k} - frac{1}{k 1}right) frac{m 1}{m 2}]

This completes the induction step.

Conclusion

We have shown through mathematical induction that the statement (sum_{k1}^n left(frac{1}{k} - frac{1}{k 1}right) frac{n}{n 1}) holds true for all positive integers (n). This is achieved by confirming the base case and the induction step, demonstrating that the series indeed simplifies to (frac{n}{n 1}) as the first and last terms cancel out in the series.

Final Note

Although the base case and the induction step confirm the result, it's important to note that this is a telescoping series, where the terms cancel out. This simplification makes it possible to evaluate the sum without extensive calculation, leading to the elegant result (frac{n}{n 1}).