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(a) Starting with the geometric series \(\Sigma_{n=0}^{\infty} X^{n},\) find the sum of the series $$\sum_{n=1}^{\infty} n x^{n-1} \quad|x|<1$$ (b) Find the sum of each of the following series. (i) $$\sum_{n=1}^{\infty} n x^{n}, \quad|x|<1 \quad\( (ii) \)\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$ (c) Find the sum of each of the following series. (i) $$\sum_{n-2} n(n-1) x^{n}, \quad|x|<1\( (ii) \)\sum_{n=2}^{\infty} \frac{n^{2}-n}{2^{n}} \quad\( (iii) \)\sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}}$$

Short Answer

Expert verified
(a) \( \frac{1}{(1-x)^2} \); (b)(i) \( \frac{x}{(1-x)^2} \), (ii) 2; (c)(i) \( \frac{2x^2}{(1-x)^3} \), (ii) 1, (iii) 6.

Step by step solution

01

Differentiate Geometric Series

Start with the geometric series \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \), for \(|x|<1\). Differentiate both sides with respect to \(x\) to find \( \sum_{n=1}^{\infty} n x^{n-1} \).The differentiation of the right-hand side gives \( \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} \), and of the left-hand side gives \( \sum_{n=1}^{\infty} n x^{n-1} \).Therefore, \( \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2} \).
02

Multiply by x for Part (b)(i)

To find the series \( \sum_{n=1}^{\infty} n x^n \), multiply the result from Step 1 by \( x \).Thus, \( x \sum_{n=1}^{\infty} n x^{n-1} = x \cdot \frac{1}{(1-x)^2} = \frac{x}{(1-x)^2} \).So, the sum of the series \( \sum_{n=1}^{\infty} n x^n \) is \( \frac{x}{(1-x)^2} \).
03

Calculate Specific Sum for Part (b)(ii)

For the series \( \sum_{n=1}^{\infty} \frac{n}{2^n} \), use \( x = \frac{1}{2} \) in the formula \( \sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} \).Substitute \( x \rightarrow \frac{1}{2} \) into \( \frac{x}{(1-x)^2} \) to get \( \frac{1/2}{(1-1/2)^2} = 2 \).
04

Second Differentiation for Part (c)(i)

Start from the formula from Step 1 and differentiate it again to find \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} \).Differentiating \( \frac{1}{(1-x)^2} \) with respect to \( x \) gives \( \frac{2}{(1-x)^3} \).Thus, \( \sum_{n=2}^{\infty} n(n-1) x^{n-2} = \frac{2}{(1-x)^3} \). Multiply by \( x^2 \) to adjust the index, thus \( \sum_{n=2}^{\infty} n(n-1) x^n = \frac{2x^2}{(1-x)^3} \).
05

Specific Sum for Part (c)(ii)

For the series \( \sum_{n=2}^{\infty} \frac{n^2-n}{2^n} \), use the result from Step 4 with \( x = \frac{1}{2} \).Substitute \( x \rightarrow \frac{1}{2} \) in \( \frac{2x^2}{(1-x)^3} \) and calculate: \( \frac{2(1/4)}{(1-1/2)^3} = 1 \).
06

Sum for squared terms in Part (c)(iii)

For \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \), use the results from Steps 1 and 4.The given series \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \) can be calculated by finding the series for \( \sum_{n=1}^{\infty} n x^n \) and then using the result of the second differentiation, similar approach gives:Consider \( x = \frac{1}{2} \), hence \( 2 + 1 = 3 \). Thus, the sum is \( 6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Summation
Series summation involves finding the total sum of all terms in a series. In the context of a geometric series, this process becomes easier thanks to specific mathematical patterns. A geometric series has the form \[S = a + ar + ar^2 + ar^3 + \cdots \]where \( a \) is the first term and \( r \) is the common ratio. For infinite geometric series with \( |r| < 1 \), the series sum can be determined by the formula:\[\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}.\]
In our exercise, we start with a special form of geometric series where each term involves a power of \( x \). Differentiating it allows us to work with sums containing terms like \( nx^n \), utilizing the differentiation technique of breaking down an infinite geometric progression.

By differentiating the simple geometric series \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \), we get new results which help us solve more complex series sums. The practicality of being able to transform a series to match known formulas is a big stepping stone in solving series summation problems.
Differentiation in Calculus
Differentiation is a powerful tool in calculus that helps us find the rate of change of a function. In the context of series, it can assist in deriving new series formulations.

For example, differentiating the geometric series can transform the series into a form that is more manageable for solving certain sums. When you differentiate the summed form of \[\sum_{n=0}^{\infty} x^n = \frac{1}{1-x},\]
you need to apply the calculus rules for differentiation:
  • The derivative of \( x^n \) becomes \( n x^{n-1} \).
  • The derivative of \( \frac{1}{1-x} \) gives \( \frac{1}{(1-x)^2} \), reflecting changes in the function's growth rate.

These operations allow us to derive the sum of the series \( \sum_{n=1}^{\infty} n x^{n-1} \). Further differentiations, as seen in the exercise, can address higher order terms like \( n(n-1) x^{n-2} \). Differentiation thus unlocks the potential to manipulate series into usable forms for direct summation or other calculations.
Convergence of Series
Convergence refers to whether the sum of a series approaches a finite number as the number of terms increases infinitely. For a geometric series, convergence occurs when the absolute value of the common ratio \( |r| < 1 \).

In our exercise, the series converge because they all maintain \(|x| < 1\). Convergence is necessary to ensure our manipulations (such as differentiation) provide valid results. Without convergence, even simple series would diverge, meaning they wouldn't sum to any meaningful value.
  • The geometric series itself converges due to the ratio condition, \(|x|<1\).
  • The further transformations by differentiation also maintain convergence due to this condition.

Understanding convergence helps verify that the sum of an altered series, such as one involving terms like \( n x^n \) or \( \frac{n}{2^n} \), indeed results in a finite and predictable value. This concept is integral to confidently manipulating series for practical solutions in calculus and mathematical analysis.

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