Problem 22
Question
A series \(\sum_{n=1}^{\infty} a_{n}\) is given. Calculate the first five partial sums of the series. That is, calculate \(S_{N}=\sum_{n=1}^{N} a_{n}\) for \(N=1,2,3,4,5\). $$ \sum_{n=1}^{\infty} 2^{n} / n ! $$
Step-by-Step Solution
Verified Answer
The first five partial sums are 2, 4, 4.67, 6 and 6.267.
1Step 1: Define the General Term
The given series is \( \sum_{n=1}^{\infty} \frac{2^n}{n!} \). The general term \( a_n \) of the series is \( \frac{2^n}{n!} \). This will be used to calculate each term individually.
2Step 2: Calculate the First Partial Sum
Calculate \( S_1 \), which is the sum of the first term.\[S_1 = a_1 = \frac{2^1}{1!} = 2.\]
3Step 3: Calculate the Second Partial Sum
Calculate \( S_2 \), which is the sum of the first and second terms.\[S_2 = a_1 + a_2 = \frac{2^1}{1!} + \frac{2^2}{2!} = 2 + 2 = 4.\]
4Step 4: Calculate the Third Partial Sum
Calculate \( S_3 \), which is the sum of the first three terms.\[S_3 = a_1 + a_2 + a_3 = \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} = 2 + 2 + \frac{8}{6} = 2 + 2 + \frac{4}{3} = \frac{14}{3} \approx 4.67.\]
5Step 5: Calculate the Fourth Partial Sum
Calculate \( S_4 \), which is the sum of the first four terms.\[S_4 = a_1 + a_2 + a_3 + a_4 = 2 + 2 + \frac{8}{6} + \frac{16}{24} = 4 + \frac{4}{3} + \frac{2}{3} = 4 + 2 = 6.\]
6Step 6: Calculate the Fifth Partial Sum
Calculate \( S_5 \), which is the sum of the first five terms.\[S_5 = a_1 + a_2 + a_3 + a_4 + a_5 = 6 + \frac{32}{120} = 6 + \frac{4}{15} = 6.267.\]
Key Concepts
Series ConvergenceFactorialsExponential Series
Series Convergence
When we talk about the convergence of a series, what we're exploring is whether the sum of an infinite series approaches a specific number as the number of terms grows. A series is said to converge if its partial sums, as we extend them to infinity, get closer to a finite limit. If the series does not converge, we say it diverges.
In the case of the given series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\), we evaluate a series of partial sums to observe the trend. Starting with small partial sums, such as \(S_1\) to \(S_5\), helps illustrate this process. As each of these partial sums is computed, if the series converges, their sum approaches a limit instead of growing without bound.
Utilizing the concept of convergence can help in verifying the behavior of infinite series and whether they are finite or infinite in sum, which is crucial in many areas of mathematics and physics.
In the case of the given series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\), we evaluate a series of partial sums to observe the trend. Starting with small partial sums, such as \(S_1\) to \(S_5\), helps illustrate this process. As each of these partial sums is computed, if the series converges, their sum approaches a limit instead of growing without bound.
Utilizing the concept of convergence can help in verifying the behavior of infinite series and whether they are finite or infinite in sum, which is crucial in many areas of mathematics and physics.
Factorials
A factorial, denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). They occur frequently in mathematics, particularly in series and combinatorics, because they help describe permutations and combinations. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow very fast with increasing \(n\).
In the context of our series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\), each term of the series involves \(n!\) in the denominator. This factorial growth in the denominator is a key factor in the decreasing size of subsequent terms, which contributes to the potential convergence of the series.
Understanding factorials helps us grasp why as \(n\) increases, the terms \(\frac{2^n}{n!}\) become smaller, leading to a converging series despite the powers of two increasing in the numerator.
In the context of our series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\), each term of the series involves \(n!\) in the denominator. This factorial growth in the denominator is a key factor in the decreasing size of subsequent terms, which contributes to the potential convergence of the series.
Understanding factorials helps us grasp why as \(n\) increases, the terms \(\frac{2^n}{n!}\) become smaller, leading to a converging series despite the powers of two increasing in the numerator.
Exponential Series
Exponential series are a special type of series that help us understand how exponential functions can be expressed as infinite sums. These series are immensely useful in both pure and applied mathematics, providing approximations and insights across science and engineering.
The series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\) represents an exponential series, notably similar to the expansion of \(e^x\) but with \(x = 2\). In the general form for the exponential function \(e^x\), the expansion is given by:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
By exploring \(2^n/n!\), we're tapping into this concept. These series can approximate exponential functions and are fundamental to calculus, particularly in solving differential equations and in modeling growth patterns.
As \(n\) grows, the terms in the exponential series become smaller, contributing to the convergence we've discussed. This makes them ideal for scenarios where precision and infinite summation are vital.
The series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\) represents an exponential series, notably similar to the expansion of \(e^x\) but with \(x = 2\). In the general form for the exponential function \(e^x\), the expansion is given by:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
By exploring \(2^n/n!\), we're tapping into this concept. These series can approximate exponential functions and are fundamental to calculus, particularly in solving differential equations and in modeling growth patterns.
As \(n\) grows, the terms in the exponential series become smaller, contributing to the convergence we've discussed. This makes them ideal for scenarios where precision and infinite summation are vital.
Other exercises in this chapter
Problem 22
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In each of Exercises 23-34, derive the Maclaurin series of the given function \(f(x)\) by using a known Maclaurin series. $$ f(x)=\sin (2 x) $$
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