Question

Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways as outlined in parts (a) and (b). a. Evaluate \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) using a telescoping series argument. b. Evaluate \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) using a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\) by obtaining a common denominator.

Short answer

Answer: The sum of the series is \(\frac{1}{2}\).

Step-by-step solution

(Evaluate using Telescoping Series Argument)

Using a telescoping series argument, we will expand the given series and look for patterns that might simplify the expression. The first few terms of the series are: \(k = 1: \frac{1}{2^1} - \frac{1}{2^2} = \frac{1}{2} - \frac{1}{4}\) \(k = 2: \frac{1}{2^2} - \frac{1}{2^3} = \frac{1}{4} - \frac{1}{8}\) \(k = 3: \frac{1}{2^3} - \frac{1}{2^4} = \frac{1}{8} - \frac{1}{16}\) Then the series becomes: \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right) = \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{16}\right) + \cdots\) Now, we can observe that the series is telescopic. Every term after the first has both a positive and negative part that is included in different elements of the sum: \(= \frac{1}{2} - \cancel{\frac{1}{4}} + \cancel{\frac{1}{4}} - \cancel{\frac{1}{8}} + \cancel{\frac{1}{8}} - \cancel{\frac{1}{16}} + \cdots\) Therefore, the series simplifies to: \(= \frac{1}{2}\)

(Evaluate using Geometric Series Argument)

In order to evaluate the series using a geometric series argument, first, we have to simplify the given expression by obtaining a common denominator. \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}} = \frac{2}{2^{k+1}}-\frac{1}{2^{k+1}} = \frac{1}{2^{k+1}}\) Now the series becomes: \(\sum_{k=1}^{\infty}\frac{1}{2^{k+1}}\) This is a geometric series with the first term \(a = \frac{1}{2^2} = \frac{1}{4}\) and the common ratio \(r = \frac{1}{2}\). To evaluate the sum, we will use the formula for the sum of an infinite geometric series: \(S = \frac{a}{1-r}\) where \(a\) is the first term, and \(r\) is the common ratio. \(S = \frac{\frac{1}{4}}{1-\frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}\) Both methods of evaluation produce the same result, so the sum of the series is \(\frac{1}{2}\).

Key concepts

Telescoping Series

When we talk about a telescoping series, we're looking at a series where most of the terms get canceled out. This type of series can often be simplified because the consecutive terms cancel each other. For instance, in the exercise provided, you initially see terms like

• \(rac{1}{2} - \frac{1}{4}\)
• \(rac{1}{4} - \frac{1}{8}\)
• \(rac{1}{8} - \frac{1}{16}\)

When written out, it becomes apparent that each negative term cancels with the next positive term. You end up with most terms vanishing except for the very first positive term. This simplification technique is remarkably useful because it often turns a potentially complicated problem into something simple and elegant, concluding that the sum of this series is \(\frac{1}{2}\).

Geometric Series

A geometric series is a series where each term is a fixed multiple of the previous term. The structure of such a series is characterized by the first term and the common ratio. In the problem, we transformed the expression into a simpler form where each term's ratio is constant. Remember, the key formula for the sum of an infinite geometric series is

• If the first term is \(a\) and the common ratio is \(r\), the sum \(S\) is \( \frac{a}{1-r}\) provided \(|r| < 1\).

In this exercise, after simplification, we got the geometric series with the first term \(a = \frac{1}{4}\) and the common ratio \(r = \frac{1}{2}\). This means the series converges and sums up to \(\frac{1}{2}\). Geometric series are particularly nice because with a given \(a\) and \(r\), you can find the sum easily and systematically.

Infinite Series

An infinite series continues indefinitely and can sometimes be tricky. The question is whether or not the series converges to a specific value.
In the case of an infinite geometric series like the one presented here, it converges due to the common ratio \(|r| < 1\).
This is essential to know because not all infinite series will converge; this depends heavily on the values and structure of the series.
Understanding whether a series converges is crucial when evaluating the sum of an infinite series. When it does converge, it means that even though it goes on forever, the series has a sum that can be found using specific mathematical techniques.

Common Denominator

Finding a common denominator is a critical step in simplifying complex fractions. It allows us to combine fractions into a single fractional expression.
In the exercise, the terms \(\frac{1}{2^{k}}\)and\(\frac{1}{2^{k+1}}\)were simplified to\(\frac{1}{2^{k+1}}\)by rewriting\(\frac{1}{2^{k}}\)as\(\frac{2}{2^{k+1}}\).This simplification was achieved by recognizing that both terms were powers of 2, thus sharing a related denominator.

• This simplification reduces the complexity in fractions, helping easily transition from fractions into forms suited for series evaluations.

Without finding a common denominator first, it would be quite challenging to proceed with the series evaluation methods effectively.