(1-1/n)+(1-2/n)+(1-3/n)+...upto N Terms

4 min read Jun 16, 2024
(1-1/n)+(1-2/n)+(1-3/n)+...upto N Terms

Understanding the Series: (1-1/n)+(1-2/n)+(1-3/n)+... upto n terms

This article delves into the fascinating world of series and aims to explore the sum of the series (1-1/n)+(1-2/n)+(1-3/n)+... upto n terms. We'll break down the steps to find the sum and provide a clear explanation for understanding the underlying concepts.

1. Identifying the Pattern

The first step is to identify the pattern present in the series. Each term can be represented as:

(1 - k/n), where k is the term number (starting from 1) and n is the total number of terms.

2. Expanding the Series

Let's expand the series to visualize the pattern more clearly:

(1 - 1/n) + (1 - 2/n) + (1 - 3/n) + ... + (1 - n/n)

3. Rearranging and Simplifying

Now, we can rearrange the terms to group the constants and the fractions:

(1 + 1 + 1 + ... + 1) - (1/n + 2/n + 3/n + ... + n/n)

This simplifies to:

n - (1/n + 2/n + 3/n + ... + n/n)

4. Finding the Sum of the Fractions

We can factor out 1/n from the second part:

n - (1/n) * (1 + 2 + 3 + ... + n)

The expression (1 + 2 + 3 + ... + n) represents the sum of the first n natural numbers. This is a well-known arithmetic series with the formula:

n(n+1)/2

5. Final Calculation

Substituting the sum of the natural numbers back into our equation, we get:

n - (1/n) * [n(n+1)/2]

Simplifying further:

n - (n+1)/2

Therefore, the sum of the series (1-1/n)+(1-2/n)+(1-3/n)+... upto n terms is (n - (n+1)/2).

Example

Let's consider an example with n = 5:

(1-1/5) + (1-2/5) + (1-3/5) + (1-4/5) + (1-5/5)

Using the formula, the sum would be:

5 - (5+1)/2 = 5 - 3 = 2

By manually calculating each term and adding them, we can verify that the sum indeed equals 2.

Conclusion

This exploration demonstrates how to find the sum of the series (1-1/n)+(1-2/n)+(1-3/n)+... upto n terms. By identifying patterns, rearranging terms, and utilizing the sum of arithmetic series, we arrived at a concise formula to calculate the sum for any given value of n. This approach highlights the power of pattern recognition and algebraic manipulation in solving series problems.

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