Problem 27 - Calculus

Question

$$\text { Find the sum of } \sum_{p=1}^{\infty} p x^{p-1} \text { for }|x|<1$$

Step-by-Step Solution

Answer

The series sums to \( \frac{1}{(1-x)^2} \) for \(|x|<1\).

Step 1: Understanding the Series

We need to find the sum of the series \( \sum_{p=1}^{\infty} p x^{p-1} \), which is an infinite series. The index \(p\) starts at 1 and ends at infinity, and each term is of the form \( p x^{p-1} \).

Step 2: Identifying Series Relation

Recognize that the given series is related to the geometric series. Recall that the sum of the geometric series \( \sum_{p=0}^{\infty} x^p \) is \( \frac{1}{1-x} \) for \(|x|<1\). The series given involves differentiation with respect to \(x\).

Step 3: Differentiating the Geometric Series

Differentiate the sum of the geometric series \( S(x) = \sum_{p=0}^{\infty} x^p = \frac{1}{1-x} \) with respect to \(x\):\[\frac{d}{dx}S(x) = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \left(\frac{1}{(1-x)^2}\right)\]

Step 4: Expressing the Desired Series

Realize that the differentiation gives us \( \sum_{p=1}^{\infty} p x^{p-1}\) by multiplying the result of differentiation by \(x\):\[x \frac{1}{(1-x)^2} = \sum_{p=1}^{\infty} p x^{p-1}\]

Step 5: Writing Final Answer

The sum of the series \( \sum_{p=1}^{\infty} p x^{p-1} \) for \(|x|<1\) is:\[\sum_{p=1}^{\infty} p x^{p-1} = \frac{1}{(1-x)^2}\] for \(|x|<1\).

Key Concepts

Geometric Series

A geometric series is one of the simplest forms of an infinite series, where each term is a constant multiple of the previous one. The standard form for a geometric series is

• \( a + ar + ar^2 + ar^3 + \ldots \)

where \( a \) is the first term, and \( r \) is the common ratio.
For the infinite geometric series with initial term 1 and a common ratio \( x \), the sum is given by \[\sum_{p=0}^{\infty}x^p = \frac{1}{1-x}\]when \( |x| < 1 \). This formula is fundamental when extending the concept to more complex series involving products of terms or incorporated within another function.
Understanding geometric series is crucial because it not only helps in finding sums but also in manipulating expressions to solve a variety of mathematical problems.

Differentiation

Differentiation is a key mathematical concept used to determine how a function changes as its input changes. In the context of the series

• \( \sum_{p=1}^{\infty} p x^{p-1}\)

we leverage differentiation on the geometric series to derive this relationship:
Differentiating the geometric sum \( S(x) = \frac{1}{1-x} \) with respect to \( x \) involves applying the chain rule. Here,

• \[\frac{d}{dx}S(x) = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}\]

By differentiating, we essentially transform the function and relate the original geometric series to our series of interest.
This application of differentiation highlights how calculus techniques can extend the utility of basic series formulas.

Sum of Series

Finding the sum of an infinite series involves understanding its limit behavior as the number of terms grows. In our exercise, we aimed to find the sum of

• \( \sum_{p=1}^{\infty} p x^{p-1}\)

The solution approached this by differentiating the sum of a geometric series and manipulating the results. This iteration leads to the result:
\[ x \frac{1}{(1-x)^2} \]
which sums intuitively to

• \( \frac{1}{(1-x)^2} \)

for \( |x| < 1 \).
Effectively, we modify the basic geometric series through calculus to obtain a precise formula for the modified series. This technique reduces the complexity of direct summation and allows for a neat, closed formula for an otherwise complex infinite process.

Convergence

Convergence in the context of infinite series refers to whether the series approaches a finite limit as its terms continue indefinitely. For a series to be useful, particularly in practical applications, it generally needs to converge. The geometric series

• \( \sum_{p=0}^{\infty}x^p \)

converges to
\[ \frac{1}{1-x} \]
only if \( |x| < 1 \). This condition ensures each additional term contributes a diminishingly small amount to the sum.
Similarly, the series in our exploration,

• \( \sum_{p=1}^{\infty} p x^{p-1} \)

also converges under \( |x| < 1 \),which is verified by its derived sum formula
\[ \frac{1}{(1-x)^2} \].
Convergence analysis is critical when dealing with infinite sums, allowing us to predict whether the series translates to a meaningful, finite result that can be applied in theories and computations.