Problem 23
Question
Transform the first expression into the second by a change of indices. $$\sum_{k=4}^{25} \frac{1}{k^{2}-9} ; \quad \sum_{n=7}^{28} \frac{1}{n^{2}-6 n}$$
Step-by-Step Solution
Verified Answer
To transform the first expression into the second, we find the relationship between the indices \(k\) and \(n\) by substituting \(n = k + 3\). Then, we can substitute and simplify the expression inside the summation: \(\frac{1}{(n - 3)^{2} - 9} = \frac{1}{n^{2} - 6n}\). Thus, the transformed summation is:
$$\sum_{n=7}^{28} \frac{1}{n^{2}-6n}$$
1Step 1: Identify relationship between k and n
We are given two summations:
$$\sum_{k=4}^{25} \frac{1}{k^{2}-9} \quad \text{and} \quad \sum_{n=7}^{28} \frac{1}{n^{2}-6n}$$
To find the relationship between the indices \(k\) and \(n\), we need to first find the starting values:
When \(k = 4\), we get \(k^2 - 9 = 7\). We want \(n^2 - 6n = 7\). Solving for \(n\), we get \(n = 7\).
2Step 2: Find the expression relationship
Since the starting value of \(n\) is 7, we can assume \(n = k + 3\). Then, the expression for \(k\) in terms of \(n\) is: \(k = n - 3\). Substitute this expression for \(k\) into the original summation expression, we get:
$$\sum_{n = 7}^{28} \frac{1}{(n - 3)^{2} - 9}$$
3Step 3: Simplify the expression
Now, we simplify the expression in the summation:
$$\frac{1}{(n - 3)^{2} - 9} = \frac{1}{n^{2} - 6n + 9 - 9}$$
So, the transformed summation becomes:
$$\sum_{n=7}^{28} \frac{1}{n^{2}-6n}$$
The first expression has now been transformed successfully into the second.
Key Concepts
Index of SummationAlgebraic ManipulationInfinite SeriesSummation Notation
Index of Summation
When dealing with a mathematical summation, it is essential to understand the role of the index of summation. This is the variable that represents each term in the series, and it changes value from one term to the next. In the context of our exercise, this index changes according to a specific rule which aligns one summation with another.
For instance, in our exercise, the transformation of the index from k in the first series to n in the second series is not arbitrary. It requires an algebraic manipulation to determine how these two indices relate to each other. When k = 4, we find that n = 7 by solving the equation for the given term's value. This jump from 4 to 7 is the starting point to define the relationship between the two indices allowing us to transform the entire summation.
For instance, in our exercise, the transformation of the index from k in the first series to n in the second series is not arbitrary. It requires an algebraic manipulation to determine how these two indices relate to each other. When k = 4, we find that n = 7 by solving the equation for the given term's value. This jump from 4 to 7 is the starting point to define the relationship between the two indices allowing us to transform the entire summation.
Algebraic Manipulation
To convert one expression into another, algebraic manipulation is often required. This term refers to the process of rearranging, combining, and simplifying algebraic expressions using the laws of algebra.
In the provided exercise, algebraic manipulation is applied by expressing the index k in terms of the index n to match the expressions in the two sums. By exploring the relationship between k and n, which in this case is n = k + 3, we can substitute k with n - 3 to transform the first summation into the form of the second. This highlights the importance of understanding and applying algebraic principles to manipulate and transform mathematical expressions effectively.
In the provided exercise, algebraic manipulation is applied by expressing the index k in terms of the index n to match the expressions in the two sums. By exploring the relationship between k and n, which in this case is n = k + 3, we can substitute k with n - 3 to transform the first summation into the form of the second. This highlights the importance of understanding and applying algebraic principles to manipulate and transform mathematical expressions effectively.
Infinite Series
While not directly the focus of this exercise, the concept of an infinite series is closely related to our summation topic. An infinite series is a summation where the number of terms is infinite, denoting a progression that goes on forever. Although the series we are dealing with in our exercise are not infinite, understanding infinite series can give you a deeper insight into the behavior of summations over large numbers of terms.
In practice, we might encounter an infinite series that converges, meaning the sum approaches a specific value, or diverges, where the sum does not settle at a single value. Summations of infinite series can be much more complex to manipulate due to their nature, but the foundational principles of handling the indexes and expressions remain the same.
In practice, we might encounter an infinite series that converges, meaning the sum approaches a specific value, or diverges, where the sum does not settle at a single value. Summations of infinite series can be much more complex to manipulate due to their nature, but the foundational principles of handling the indexes and expressions remain the same.
Summation Notation
Finally, to make sense of these expressions, it is crucial to be familiar with summation notation. This notation is employed to concise and clearly communicate the summation of a sequence of terms. It uses the Greek letter Sigma (\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f)), metavariables, an expression to be summed, and limits to indicate where the summation starts and ends.
In our exercise, the term \f$$\f(\f($$$f(\frac{1}{k^2-9}sum\frac{1}{k^2-9}sum $$\f)sum gives a clear instruction on what operation to perform: starting with \f\(k = 4sum \)\f)sum$$\f)sum$$\f)sum\(\f(\frac{1}{k^2-9}$$\f)sumk\)\f(\frac{1}{k-3})$$\f)\f(\frac{1}{k-3})sumkf \(\f(\frac{1}{k-3})$$f $$\f(\frac{1}{k-3})sum \)\f(\frac{1}{k-3})\f$$\f(\frac{1}{k-3})sum\frac{1}{k-3}sum_{\(f{\frac{1}{k-3}sum_{\)f\frac{1}{k-3}sum_{\(f$$\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f(\f(\f)\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f)\f(\f(\f)$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f-237 Summit-237 Summit, all the way up to \)n = 28sum of_sum $traversal. This economical tool greatly simplifies the communication of complex sum operations and serves as a universal language among mathematicians and students alike.
In our exercise, the term \f$$\f(\f($$$f(\frac{1}{k^2-9}sum\frac{1}{k^2-9}sum $$\f)sum gives a clear instruction on what operation to perform: starting with \f\(k = 4sum \)\f)sum$$\f)sum$$\f)sum\(\f(\frac{1}{k^2-9}$$\f)sumk\)\f(\frac{1}{k-3})$$\f)\f(\frac{1}{k-3})sumkf \(\f(\frac{1}{k-3})$$f $$\f(\frac{1}{k-3})sum \)\f(\frac{1}{k-3})\f$$\f(\frac{1}{k-3})sum\frac{1}{k-3}sum_{\(f{\frac{1}{k-3}sum_{\)f\frac{1}{k-3}sum_{\(f$$\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f)\f(\f(\f(\f)\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f)\f(\f(\f)$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f$$f-237 Summit-237 Summit, all the way up to \)n = 28sum of_sum $traversal. This economical tool greatly simplifies the communication of complex sum operations and serves as a universal language among mathematicians and students alike.
Other exercises in this chapter
Problem 23
Find the interval of convergence. $$\sum(-1)^{k} \frac{2^{k}}{3^{k+1}} x^{k}$$
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Test these series for (a) absolute convergence, (b) conditional convergence. \(\sum(-1)^{k} k \sin (1 / k)\).
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Assume that a ball dropped to the floor rebounds to a height proportional to the height from which it was dropped. Find the total length of the path of a ball d
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Determine whether the series converges or diverges. $$\sum k^{-(1+1 / k)}$$
View solution