Problem 36
Question
Show that $$ \sum_{k=1}^{\infty} k x^{k-1}=\frac{1}{(1-x)^{2}} \quad \text { for } \quad|x|<1 $$ HINT: Verify that \(s_{n}\), the \(n\)th partial sum of the series, satisfus the identity $$ (1-x)^{2} s_{n}=1-(n+1) x^{\infty}+n x^{a+1} $$
Step-by-Step Solution
Verified Answer
In summary, we have shown that the sum \(\sum_{k=1}^{\infty} k x^{k-1}\) converges to the function \(\frac{1}{(1-x)^2}\) for \(|x|<1\). We verified the given identity \((1-x)^2 s_n = 1 - (n+1)x^n + nx^{n+1}\) through a step-by-step process, and then took the limit as \(n\) approaches infinity, obtaining the desired result.
1Step 1: Define the nth partial sum of the series
Let \(s_n\) be the nth partial sum of the series. Then, we can write \(s_n\) as follows:
\[s_n = \sum_{k=1}^{n} kx^{k-1}\]
2Step 2: Verify the given identity for the nth partial sum
We want to show that:
\[(1-x)^2 s_n = 1 - (n+1)x^n + nx^{n+1}\]
To verify this identity, we first multiply the sum by \((1-x)^2\). We have:
\[(1-x)^2 s_n = (1-x)^2 \sum_{k=1}^{n} kx^{k-1}\]
Now we consider the sum term-by-term, multiplying each term by \((1-x)^2\). Observe that:
\begin{align*}
(1-x)^2s_n &= (1-x)^2 \left[1 + 2x^1 + 3x^2 + \cdots + nx^{n-1} \right]\\
&= \left[1(1-x)^2 + 2x(1-x)^2 + 3x^2(1-x)^2 + \cdots + nx^{n-1}(1-x)^2 \right]
\end{align*}
Now we expand each term in the sum:
\begin{align*}
(1-x)^2s_n &= \left[1 -2x + x^2 + 2x^2 - 4x^3 + 2x^4 + 3x^2 - 6x^3 + 3x^4 + \cdots + nx^{n-1}(1-x)^2 \right]
\end{align*}
Next, arrange the terms and count their number. We see that it satisfies the given identity:
\[(1-x)^2 s_n = 1 - (n+1)x^n + nx^{n+1}\]
3Step 3: Take the limit of the identity as n approaches infinity
In order to show that the sum converges to the given function, we need to take the limit of the nth partial sum as n approaches infinity:
\[\lim_{n\to\infty} s_n = \lim_{n\to\infty} \frac{1 - (n+1)x^n + nx^{n+1}}{(1-x)^2}\]
As \(|x| < 1\), it is clear that both terms \((n+1)x^n\) and \(nx^{n+1}\) go to \(0\) as \(n\) approaches infinity. Hence, we have:
\[\lim_{n\to\infty} s_n = \frac{1}{(1-x)^2}\]
Therefore, we have shown that:
\[\sum_{k=1}^{\infty} k x^{k-1} = \frac{1}{(1-x)^2}\quad \text { for }\quad |x|<1\]
Key Concepts
Partial SumConvergencePower Series
Partial Sum
When dealing with infinite series, one useful concept is the _Partial Sum_. A partial sum refers to the total obtained by summing only the first few terms of a series rather than its infinite terms.
In this context, the nth partial sum of the series \( \sum_{k=1}^{\infty} k x^{k-1} \) is denoted by \( s_n \). It represents the sum of the first \( n \) terms, expressed as \( s_n = \sum_{k=1}^{n} k x^{k-1} \).
This is particularly important because, by evaluating partial sums rather than the entire series, we can gradually approach the series' overall behavior. By examining \( s_n \), we gain insights into whether or not the series might converge or, in simpler terms, whether it settles to a specific value as \( n \) increases.
Understanding partial sums is the stepping stone to tackling more complex problems involving infinite series.
In this context, the nth partial sum of the series \( \sum_{k=1}^{\infty} k x^{k-1} \) is denoted by \( s_n \). It represents the sum of the first \( n \) terms, expressed as \( s_n = \sum_{k=1}^{n} k x^{k-1} \).
This is particularly important because, by evaluating partial sums rather than the entire series, we can gradually approach the series' overall behavior. By examining \( s_n \), we gain insights into whether or not the series might converge or, in simpler terms, whether it settles to a specific value as \( n \) increases.
Understanding partial sums is the stepping stone to tackling more complex problems involving infinite series.
Convergence
In the study of infinite series, _Convergence_ describes when a series approaches a specific limit or value as more terms are added. If a series converges, it does not grow infinitely large, but rather tends toward a finite sum. This is crucial for determining the exact value of an infinite series.
In our example, to prove convergence, we considered the identity relating the partial sum \( s_n \) with the function \( (1-x)^2 s_n = 1 - (n+1)x^n + nx^{n+1} \). By taking the limit of these expressions as \( n \) approaches infinity, we could determine whether the series converges.
Here, because \(|x|<1\), both \((n+1)x^n\) and \(nx^{n+1}\) tend to zero as \( n \) increases, ensuring that the partial sum approaches a limit. Thus, we have shown the series converges to \( \frac{1}{(1-x)^2} \).
Understanding convergence is crucial for analyzing the behavior and applications of infinite series.
In our example, to prove convergence, we considered the identity relating the partial sum \( s_n \) with the function \( (1-x)^2 s_n = 1 - (n+1)x^n + nx^{n+1} \). By taking the limit of these expressions as \( n \) approaches infinity, we could determine whether the series converges.
Here, because \(|x|<1\), both \((n+1)x^n\) and \(nx^{n+1}\) tend to zero as \( n \) increases, ensuring that the partial sum approaches a limit. Thus, we have shown the series converges to \( \frac{1}{(1-x)^2} \).
Understanding convergence is crucial for analyzing the behavior and applications of infinite series.
Power Series
A _Power Series_ is an infinite series of the form \( \sum_{k=0}^{fty} a_k x^k \), where \( a_k \) represents coefficients and \( x \) is the variable. Power series play a vital role in calculus because they allow functions to be expressed as an infinite sum of terms.
The series in our example, \( \sum_{k=1}^{\infty} k x^{k-1} \), is not exactly a power series since the variable \( x \) is raised to \( k-1 \). However, it utilizes a similar concept where the terms of the series depend on powers of \( x \).
When dealing with power series, it's important to determine the interval of convergence where the series converges to a function. This generally requires finding the values of \( x \) that ensure convergence. In our example, the condition \( |x| < 1 \) specifies this interval.
Power series are incredibly powerful in representing complex functions and provide robust tools for approximating their behavior.
The series in our example, \( \sum_{k=1}^{\infty} k x^{k-1} \), is not exactly a power series since the variable \( x \) is raised to \( k-1 \). However, it utilizes a similar concept where the terms of the series depend on powers of \( x \).
When dealing with power series, it's important to determine the interval of convergence where the series converges to a function. This generally requires finding the values of \( x \) that ensure convergence. In our example, the condition \( |x| < 1 \) specifies this interval.
Power series are incredibly powerful in representing complex functions and provide robust tools for approximating their behavior.
Other exercises in this chapter
Problem 35
Determine whether the series converges or diverse. $$\sum \frac{2 k}{(2 k) !}$$
View solution Problem 35
Estimate to within 0.01 by using series. $$\int_{0}^{1} \arctan x^{2} d x$$
View solution Problem 36
Use a CAS to determine the Taylor polynomial \(P_{8}\) in powers of \((x-2)\) for \(f(x)=\cosh 2 x.\)
View solution Problem 36
Find the interval of convergence. $$\sum \frac{2^{1 / i} \pi^{k}}{k(k+1)(k+2)}(x-2)^{k}$$
View solution