Question

In the event that a series converges uniformly, one can consider the derivative of the series to arrive at the summation of other infinite series. a. Differentiate the series representation for \(f(x)=\frac{1}{1-x}\) to sum the series \(\sum_{n=1}^{\infty} n x^{n},|x|<1\) b. Use the result from part a to sum the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) c. Sum the series \(\sum_{n=2}^{\infty} n(n-1) x^{n},|x|<1\) d. Use the result from part \(c\) to sum the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) e. Use the results from this problem to sum the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\)

Short answer

a) The sum of the series is \(\frac{1}{(1-x)^2}\). b) The sum is 25. c) The sum is \(\frac{2} {(1-x)^3}\). d) The sum is 200. e) The sum is 125.

Step-by-step solution

Differentiating the Function

The function \(f(x)=\frac{1}{1-x}\) can be represented as a series, \(\sum_{n=0}^{\infty} x^{n}\). If we differentiate this series term by term, the derived series is \(\sum_{n=1}^{\infty} n x^{n-1}\).

Summing the Derived Series

The series \(\sum_{n=1}^{\infty} n x^{n-1}\) can be written as \(\sum_{n=1}^{\infty} n x^{n}\), where \(|x|<1\). Thus, the sum of this series is \(\frac{1}{(1-x)^2}\).

Sum a Specific Series

For the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\), we substitute \(x=\frac{1}{5}\) into \(\frac{1}{(1-x)^2}\) to get \(\frac{1}{(1-\frac{1}{5})^2} = 25\).

Differentiating and Summing another Series

We apply the derivative a second time on the series from Step 1, resulting \(\sum_{n=2}^{\infty} n(n-1)x^{n-2}\). Rearranging, we get \(\sum_{n=2}^{\infty} n(n-1)x^{n}\), which sums up to \(\frac{2} {(1-x)^3}\), where \(|x|<1\).

Sum the Derived Series

We take the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) and put \(x=\frac{1}{5}\) into \(\frac{2} {(1-x)^3}\), yielding \(200\).

Sum the final series

To find the summation of the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\), subtract 3 times the summation of \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) (obtained in Step 3) from the summation of \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) (obtained in Step 5). So, \(200 - 3*25 = 125\).

Key concepts

Infinite Series

An infinite series is an expression that represents the sum of the terms of an infinite sequence. Imagine you are adding up numbers one after another without stopping. An infinite series can show up in the form of power series, geometric series, geometric series, etc.

For example, the function \[f(x) = \frac{1}{1-x}\]is represented as an infinite series:\[\sum_{n=0}^{\infty} x^{n}\]This representation is valid for values of \(|x| < 1\). It forms a geometric series, where each term is simply \(x\) raised to the power of the term number.

This makes it easy to work with because, with this representation, we can do things like differentiate or integrate term by term. It's especially useful in calculus when dealing with complex functions as it simplifies the problem into manageable pieces.

Differentiation of Series

Differentiating a series involves taking the derivative of each term in the series. This is an incredible tool, particularly when working with power series, because it allows us to find new series that represent derivatives of functions.

Here's how it works with our function \(f(x)=\frac{1}{1-x}\), which can be expanded as:\[\sum_{n=0}^{\infty} x^{n}.\]When you differentiate term-by-term, each term \(x^n\) becomes \(n x^{n-1}\).

Therefore, the result is:\[\sum_{n=1}^{\infty} n x^{n-1},\]dropping the \(x^0\) term that gives you \(n=0\). Replacing \(x^{n-1}\) with \(x^n\) for simplification purpose yields:\[\sum_{n=1}^{\infty} n x^{n}.\]

It's crucial when working with differentiation of series that the series converges uniformly, ensuring that we can safely perform these operations.

Convergence of Series

Convergence of a series refers to whether the series approaches a finite value as more terms are added. This concept is fundamental as it dictates when and how we can manipulate series, such as through differentiation or integration.

The convergence of a series \(\sum_{n=0}^{\infty} x^{n}\) is important because if a series doesn't converge, the sum isn't finite, and our mathematical operations wouldn't necessarily be valid.

• Uniform convergence means the series converges to the same value for all x within a specified radius, |x| < 1, in the case of \(\sum_{n=0}^{\infty} x^{n}\).

Let's illustrate convergence with the geometric series we started with. The formula evaluating this series is given by:\[\frac{1}{1-x},\]which is valid when |x| < 1.

This tells us that as you add more terms under these conditions, the series comes closer to \(\frac{1}{1-x}\), ensuring we can safely combine operations such as differentiation or integration.