Problem 60 - Thomas Calculus

Question

The sum of the series \(\sum_{n=0}^{\infty}\left(n^{2} / 2^{n}\right)\) To find the sum of this series, express 1\(/(1-x)\) as a geometric series, differentiate both sides of the resulting equation with respect to \(x,\) multiply both sides of the result by \(x\) , differentiate again, multiply by \(x\) again, and set \(x\) equal to 1\(/ 2 .\) What do you get?

Step-by-Step Solution

Answer

The sum of the series \(\sum_{n=0}^{\infty} \frac{n^2}{2^n}\) is 6.

Step 1: Expressing the Geometric Series

The geometric series can be expressed as \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) for \(|x| < 1\). This will form the basis for differentiation.

Step 2: First Differentiation

Differentiate both sides with respect to \( x \). The derivative of \( \frac{1}{1-x} \) with respect to \( x \) is \( \frac{1}{(1-x)^2} \). The derivative of \( \sum_{n=0}^{\infty} x^n \) is \( \sum_{n=1}^{\infty} n \cdot x^{n-1} \).

Step 3: Multiply by x

After differentiation, multiply both sides by \( x \). This gives \( \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n \cdot x^n \).

Step 4: Second Differentiation

Differentiate the resulting equation again with respect to \( x \). The left side becomes \( \frac{(1+x)}{(1-x)^3} \) after simplifying. The right side, \( \sum_{n=1}^{\infty} n^2 \cdot x^{n-1} \), differentiates to \( \sum_{n=1}^{\infty} n^2 \cdot x^{n-1} \).

Step 5: Multiply by x Again

Multiply the differentiated equation by \( x \) to get \( \frac{x(1+x)}{(1-x)^3} = \sum_{n=1}^{\infty} n^2 \cdot x^n \).

Step 6: Substitute x=1/2

Set \( x = \frac{1}{2} \) in the equation \( \frac{x(1+x)}{(1-x)^3} = \sum_{n=1}^{\infty} n^2 \cdot x^n \). This evaluates to \( \frac{\frac{1}{2}(1+\frac{1}{2})}{(1-\frac{1}{2})^3} = \sum_{n=1}^{\infty} n^2 \cdot \left(\frac{1}{2}\right)^n \).

Step 7: Simplify and Evaluate

Simplifying \( \frac{\frac{1}{2}(1+\frac{1}{2})}{(1-\frac{1}{2})^3} \) leads to \( \frac{\frac{1}{2} \cdot \frac{3}{2}}{\left(\frac{1}{2}\right)^3} = \frac{\frac{3}{4}}{\frac{1}{8}} = 6 \). Thus, the sum of the series \( \sum_{n=0}^{\infty} \frac{n^2}{2^n} \) is 6.

Key Concepts

Geometric Series

A geometric series is a series of numbers where each term is a constant multiple of the previous one. It looks like this:

• Start with a number (called the first term)
• Multiply it by a constant amount each time (called the common ratio)

In our exercise, the geometric series is expressed as \[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \]for \(|x| < 1\). This formula is very useful because it lets us transform complicated expressions into simpler ones. It’s important to remember that the series converges only when the absolute value of x is less than 1.
Convergence means that as you keep adding more and more terms, you end up with a really nice, fixed number. This property of geometric series is what allows us to go on and differentiate it!

Differentiation

Differentiation is a powerful tool in calculus used to compute the rate at which a quantity changes. In our step-by-step process:

• First, we differentiate both sides of the geometric series expression with respect to \( x \).
• After this, the expression becomes more complex, but importantly, the differentiation allows us to create a pattern of terms we can work with.

The first differentiation turns \( \frac{1}{1-x} \) into \( \frac{1}{(1-x)^2} \), showing how rapidly the geometric series changes at any point of \( x \). We then multiply this by \( x \), making it easier to apply the differentiation rule again.
The second differentiation further transforms \( \frac{x}{(1-x)^2} \) into \( \frac{(1+x)}{(1-x)^3} \), fetching another layer of complexity. This complexity helps in finding more specific terms of our series sum, bridging the gap between differentiation and series evaluation.

Infinite Series

An infinite series is simply a sum of infinitely many terms. Unlike finite series, you keep on adding numbers without stopping. For exercises like this:

• We look for a simple expression that shows the end result of this infinite sum.
• We determine whether the infinite series converges to a particular value.

In our exercise, we found the infinite series sum converges to 6 by:

• Evaluating the differentiated function at \( x = \frac{1}{2} \), leading to \( \sum_{n=1}^{\infty} n^2 \cdot \left(\frac{1}{2}\right)^n \)
• Finally simplifying and evaluating to find the sum is equal to 6.

Understanding infinite series is crucial when dealing with calculus and convergence because it involves the potential to add up an unlimited number of terms and still land on a single, rational number. It unlocks the capability of simplifying complex problems and predicting behavior in mathematical models.