Problem 36
Question
In Exercises \(35-38,\) write an equivalent series with the index of summation beginning at \(n=1\). $$ \sum_{n=0}^{\infty}(-1)^{n+1}(n+1) x^{n} $$
Step-by-Step Solution
Verified Answer
The series equivalent to the one given, but starting summing from \(n=1\), is \(\sum_{n=1}^{\infty}(-1)^{n}(n) x^{n-1}\).
1Step 1: Understand the given series
Here the series given is \(\sum_{n=0}^{\infty}(-1)^{n+1}(n+1) x^{n}\), which is an infinite series from \(n=0\) to \(\infty\), and each term in the series is \((-1)^{n+1}(n+1) x^{n}\).
2Step 2: Identify a transformation
To change the starting index to 1, substitute \(n = m-1\), where \(m\) is the new index. To get the same terms of the series when \(n=0\), we need to let \(m=n+1\). Consequently, when \(n=0\), \(m=1\) and when \(n=\infty\), \(m=\infty\).
3Step 3: Apply the transformation
Applying this transformation to every \(n\) in the sequence gives a new sequence \(\sum_{m=1}^{\infty}(-1)^{m}(m) x^{m-1}\), or more commonly expressed as \(\sum_{n=1}^{\infty}(-1)^{n}(n) x^{n-1}\).
Key Concepts
Series Index TransformationConvergent SeriesSummation Notation
Series Index Transformation
When working with a series, particularly those within summation notation, we sometimes encounter the need to adjust the range over which we're summing. This is known as a series index transformation, which allows us to rewrite a series starting from a different index without changing its value.
Let's demystify this concept with an example. Consider the series given in our exercise: \[\begin{equation}\sum_{n=0}^{\infty}(-1)^{n+1}(n+1) x^{n}\end{equation}\] Here, to change the starting index of summation from 0 to 1, a transformation is necessary. We approach it by redefining the index: setting the new index, say, \(m = n+1\). This effectively shifts all terms in the series 'forward' by one position.
After applying this transformation, we obtain a new equivalent series with the terms arranged starting from \(m=1\), which is:\[\begin{equation}\sum_{m=1}^{\infty}(-1)^{m}(m) x^{m-1}\end{equation}\]
Let's demystify this concept with an example. Consider the series given in our exercise: \[\begin{equation}\sum_{n=0}^{\infty}(-1)^{n+1}(n+1) x^{n}\end{equation}\] Here, to change the starting index of summation from 0 to 1, a transformation is necessary. We approach it by redefining the index: setting the new index, say, \(m = n+1\). This effectively shifts all terms in the series 'forward' by one position.
After applying this transformation, we obtain a new equivalent series with the terms arranged starting from \(m=1\), which is:\[\begin{equation}\sum_{m=1}^{\infty}(-1)^{m}(m) x^{m-1}\end{equation}\]
Why is Index Transformation Important?
- It makes it easier to compare series with similar forms but different starting indices.
- It allows the application of certain convergence tests which may require the series to start at a specific index.
- It aids in the simplification of series, especially when integrating or differentiating term-by-term.
Convergent Series
Moving on from adjusting indices, we step into the realm of determining whether a series 'adds up' to a limiting value or not - this is what forms the basis of a convergent series. A series is considered convergent if the sum of its infinite terms approaches a finite limit. If no such limit exists, the series is said to be divergent.
To clarify, let's consider an infinite series:\[\begin{equation}\sum_{n=1}^{\infty}a_n\end{equation}\]If this series adds up to a certain number, say \(L\), as \(n\) approaches infinity, then \(L\) is the limit to which the series converges.
For the series in our exercise, whether it converges or not can be examined by several tests, such as the ratio test, the root test, and the alternating series test, among others. If the series does converge, the transformed series we've computed will also converge, as it represents the same sum albeit in a re-indexed manner.
To clarify, let's consider an infinite series:\[\begin{equation}\sum_{n=1}^{\infty}a_n\end{equation}\]If this series adds up to a certain number, say \(L\), as \(n\) approaches infinity, then \(L\) is the limit to which the series converges.
For the series in our exercise, whether it converges or not can be examined by several tests, such as the ratio test, the root test, and the alternating series test, among others. If the series does converge, the transformed series we've computed will also converge, as it represents the same sum albeit in a re-indexed manner.
Summation Notation
In mathematics, summation notation, signified by the symbol \(\sum\), is a convenient way to express the addition of series of numbers. When you see\[\begin{equation}\sum_{n=1}^{k} a_n\end{equation}\]it means to sum the sequence \(a_n\) from \(n=1\) to \(n=k\). The \(n\) under the summation symbol is called the index of summation, and \(a_n\) represents the nth term of the series.
The power of summation notation lies in its ability to compactly represent very large sums. For example, instead of writing out \(a_1 + a_2 + \dots + a_k\), we can simply use the summation symbol to condense this expression, making it more manageable especially when dealing with series involving a large number of terms or complex patterns within those terms.
Understanding how to interpret and manipulate summation notation is crucial for working with series, performing calculations, and understanding the behavior of mathematical functions represented as series.
The power of summation notation lies in its ability to compactly represent very large sums. For example, instead of writing out \(a_1 + a_2 + \dots + a_k\), we can simply use the summation symbol to condense this expression, making it more manageable especially when dealing with series involving a large number of terms or complex patterns within those terms.
Understanding how to interpret and manipulate summation notation is crucial for working with series, performing calculations, and understanding the behavior of mathematical functions represented as series.
Other exercises in this chapter
Problem 36
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In Exercises \(35-38,\) use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error. $$ e \app
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Use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2}\left(e^{i x}+e^{-i x}\right)=\cos x $$
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