Question

Find the sum for each of the series: a. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) b. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). c. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). d. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)} .\)

Short answer

The sums for the series are \(\frac{12}{5}\), \(\frac{1}{10}\), \(\frac{23}{2}\), and \(\frac{11}{6}\) respectively.

Step-by-step solution

Solve the first series

The first series is given by \( \sum_{n=0}^{\infty} \frac{(-1)^n 3}{4^n} \). The sum \( S \) can be found using the formula \( \frac{a}{1 - r} \), where \( a = 3 \) and \( r = -\frac{1}{4} \), i.e. \( S = \frac{3}{1 - (-\frac{1}{4})} = \frac{3}{1 + \frac{1}{4}} = \frac{3}{\frac{5}{4}} = \frac{12}{5} \).

Solve the second series

The second series is given by \( \sum_{n=2}^{\infty} \frac{2}{5^n} \). We can redefine \( n \) such that the summation starts from 0. So, it becomes \( \sum_{n=0}^{\infty} \frac{2}{5^{n+2}} = \sum_{n=0}^{\infty} \frac{2}{25} \times \frac{1}{5^n} \). This becomes a geometric series with \( a = \frac{2}{25} \) and \( r = \frac{1}{5} \). Using the sum formula, we get \( S = \frac{a}{1 - r} = \frac{2/25}{1 - \frac{1}{5}} = \frac{2/25}{\frac{4}{5}} = \frac{2/25}{\frac{5}{4}} = \frac{2}{20} = \frac{1}{10} \).

Solve the third series

The third series is \( \sum_{n=0}^{\infty} (\frac{5}{2^n} + \frac{1}{3^n}) \). Splitting the series, we get \( \sum_{n=0}^{\infty} \frac{5}{2^n} + \sum_{n=0}^{\infty} \frac{1}{3^n} \). Solving individually, we get \( S1 = \frac{5}{1 - \frac{1}{2}} = 10 \) and \( S2 = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \). Summing these up, \( S = S1 + S2 = 10 + \frac{3}{2} = \frac{23}{2} \).

Solve the fourth series

The fourth series given is \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} \). This can be rewritten as \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} = \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+3}) \). The series telescopes and thus, the sum is \( S = 1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{6} + ... = 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} = \frac{11}{6} \).

Key concepts

Series Summation

When dealing with series, the main goal is to find the sum of infinitely many terms. In mathematics, a series is essentially a sum of a sequence of numbers. The sequence can be anything—geometric, arithmetic, or something else. Understanding series summation means understanding the process of adding these numbers to reach a finite number or determining if the series diverges.

To find the sum of a series:

• Identify the first term and common ratio/difference (if applicable).
• Use appropriate formulas for convergence.
• Simplify expressions to get the sum in an understandable form.

In many cases, knowing these steps can guide you in finding the sum of even the most complex of series.

Telescoping Series

A telescoping series is a special type of series where each term cancels out a part of another term, making it easier to find the sum. It's like watching dominos fall, where each piece leads to the collapse of another piece.

Consider the series given by \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} \), which can be rewritten using partial fractions. By writing it as \( \frac{1}{n} - \frac{1}{n+3} \), each pair of terms effectively cancels out with subsequent terms.

Key steps in solving a telescoping series:

• Rewrite the terms so they telescope.
• Notice which terms cancel out.
• Focus on the first few and the last few remaining terms to find the sum.

These simplifications often lead to quick calculations of sums.

Geometric Progression

A geometric progression is a sequence of numbers where each term after the first is found by multiplying or dividing the previous one by a fixed, non-zero number called the common ratio.

For instance, in the series \( \sum_{n=0}^{\infty} \frac{(-1)^n 3}{4^n} \), it's a geometric series with the first term \( a = 3 \) and common ratio \( r = -\frac{1}{4} \). The sum of an infinite geometric series can be calculated using: \[ S = \frac{a}{1 - r} \] provided the absolute value of \( r \) is less than 1.

Things to remember:

• If \(|r| < 1\), the series converges to a sum \( S \).
• Calculate using the formula for quick results.
• This concept simplifies solving many problems involving repeated multiplication.

This powerful concept is widely used across various real-life scenarios, from calculating interest to optimizing processes.