Activity 10.4.1.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}
\end{equation*}
Section 10.4 Week 10 Activity
Subsection 10.4.1 Alternating Series
Activity 10.4.2.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1}(n^2 + 3n + 4)}{4n^2 + 2n
+ 4}
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{n^2 + 3n + 4}{4n^2 + 2n +
4} = \frac{1}{4}
\end{equation*}
This is an asymptotic analysis limit and the order is the same in both numerator and denominator. The limit is the ratio of the leading terms. Since the limit is not zero, the series diverges by the alternating series test.
Activity 10.4.3.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} \sin ((2n+1)\pi)
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \sin ((2n+1)\pi) = \lim_{n
\rightarrow \infty} 0 = 0
\end{equation*}
This is a limit of constant zeros, so it is certainly zero, and the series converges. (Adds up to 0, in fact, since it is just a sum of zeros!)
Activity 10.4.4.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} \sqrt{n}
\end{equation*}
Solution.
Activity 10.4.5.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} \ln \left( \frac{1}{n} \right)
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \ln \left( \frac{1}{n} \right) =
-\infty
\end{equation*}
The limit of the terms is not zero, so the series is divergent.
Activity 10.4.6.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} e^{-n}
\end{equation*}
Solution.
Activity 10.4.7.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{4}{1+n}
\end{equation*}
Activity 10.4.8.
Test this series for convergence.
\begin{equation*}
\sum_{n=2}^{\infty} \frac{1}{n \ln n}
\end{equation*}
Solution.
I will use the integral test to deal with the logarithm. In the resulting integral, I use a substitution \(u = \ln
x\) with \(du = \frac{1}{x} dx\text{.}\)
\begin{equation*}
\int_2^\infty \frac{1}{x \ln x} dx = \int_{\ln 2}^\infty
\frac{1}{u} du = \lim_{a \rightarrow \infty} \ln u
\Bigg|_{\ln 2}^a = \lim_{a \rightarrow \infty} \ln a - \ln
(\ln 2) = \infty
\end{equation*}
The integral diverges, so the sum will also diverge.
Activity 10.4.9.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{1 + 2^n + 4^n}{7^n}
\end{equation*}
Solution.
I can split this up into three series using linearity.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{1}{7^n}
+ \sum_{n=1}^{\infty} \frac{2^n}{7^n}
+ \sum_{n=1}^{\infty} \frac{4^n}{7^n}
\end{equation*}
All three of these are geometric series. The common ratios are \(\frac{1}{7}\text{.}\) \(\frac{2}{7}\text{,}\) and \(\frac{4}{7}\text{.}\) All the common ratios are less than one, so all three series converge, and I conclude that the original series also converges.
Activity 10.4.10.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{\sqrt{n^6 + 7n^3 +
1}}{\sqrt[3]{n^9 + 3n}}
\end{equation*}
Solution.
I consider the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{\sqrt{n^6 + 7n^3 +
1}}{\sqrt[3]{n^9 + 3n}}
\end{equation*}
The asymptotic order of the numerator is \(\sqrt{n^6} =
n^3\text{.}\) The asymptotic order of the denominator is \(\sqrt[3]{n^9} = n^3\text{.}\) The limit here is the ratio of the leading coefficients, since the asymptotic order of the numerator is the same as the denominator. That ratio is 1. The limit of the terms is not zero, so by the test for divergence, the series diverges.
Activity 10.4.11.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{e^n}{n!}
\end{equation*}
Solution.
With factorials, I’ll use the ratio test. I take the limit of the ratio of the terms. In the limit, note how the factorials cancal: \(\frac{n!}{(n+1)!} = \frac{1}{n+1}\text{.}\)
\begin{equation*}
\lim_{n \rightarrow \infty}
\frac{\left| \frac{e^{n+1}}{(n+1)!} \right|}{\left|
\frac{e^n}{n!} \right|} = \lim_{n
\rightarrow \infty} \frac{n!}{(n+1)!} \frac{e^{n+1}}{e^n} =
\lim_{n \rightarrow \infty} \frac{e}{n+1} = 0
\end{equation*}
The outcome of the ratio test is less than one, so the limit converges.
Activity 10.4.12.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} 2^{\frac{1}{n}}
\end{equation*}
Activity 10.4.13.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{\sin n}{n^2}
\end{equation*}
Activity 10.4.14.
Test this series for convergence.
\begin{equation*}
\sum_{n=4}^{\infty} \frac{\ln (n-3)}{n-3}
\end{equation*}
Solution.
Since the logarithm is a growing function, these terms are eventually larger than \(\frac{1}{n-3}\text{.}\) Those terms are (with a shift) the terms of the divergence harmonic series. Therefore, since these terms are larger than the terms of a divergent series, this diverges by direct comparison.
Subsection 10.4.2 Conceptual Review Questions
Subsection 10.4.3 Extra Practice Activities
Activity 10.4.15.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1} (n^6 + 54}{n!}
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{n^6 + 54}{n!} = 0
\end{equation*}
Asymptotically, the factorial grows faster than any polynomial or exponential function. Therefore, the limit is zero and the series converges.
Activity 10.4.16.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} \sin \left( \frac{1}{n} \right)
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \sin \left( \frac{1}{n} \right)
= 0
\end{equation*}
The limit of the terms is zero, so the series converges. In this limit, see that the term inside the sine fucntion are getting close to zero. Since sine is continuous and \(\sin
(0) = 0\text{,}\) applying the sine function to those terms gives new terms which will still be approaching zero.
Activity 10.4.17.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty (-1)^{n+1} \cos \left( \frac{1}{n} \right)
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \cos \left( \frac{1}{n} \right)
= 1
\end{equation*}
The limit does not approach zero, so the series is divergent.
Activity 10.4.18.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1} (n!)(n+1)!}{(2n+3)!}
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms. For this limit, I have to do some factorial algebra. I’ll split up the factorial in the denominator into three pieces.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{n!(n+1)!}{2n+3)!} =
\lim_{n \rightarrow \infty} \frac{n!}{n!}
\frac{(n+1)!}{(n+1)(n+2)\ldots (2n+2)} \frac{1}{(2n+3)}
\end{equation*}
Let me look at these three pieces. In the first, piece, the \(n!\) cancels off, leaving just \(1\text{.}\) The third piece is simply \(\frac{1}{2n+3}\text{.}\) The middle piece is the most interesting. Here there are \(n+1\) numers in the numerator multiplied together and there are also \(n+1\) numbers in the denominator multiplied together. In the numerator, the numbers multiplied at all numbers up to \(n+1\text{.}\) In the denominator, the numbers multiplied together are the numbers starting with \(n+1\) and getting larger. Therefore, the product in the denominator must be larger than the product in the numerator and the fraction must be less than one. Putting it all together, I get constnat 1 multiplies by something less than one multiplied by \(\frac{1}{2n+3}\text{,}\) which approach zero. Therefore, the limit must be zero as well, which means that the series converges.
Activity 10.4.19.
Use the alternating series to test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{1 - \frac{1}{n}}
\end{equation*}
Solution.
The alternating series test says that an alternating series converges if and only if the limit of the terms is zero. I calculate the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{1}{1 - \frac{1}{n}} = 1
\end{equation*}
The limit of the terms is not zero, so the sequence is divergent.
Activity 10.4.20.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{(n+5)!}{(n+7)!}
\end{equation*}
Solution.
I can simplify the terms by canceling with the factorial.
\begin{equation*}
\frac{(n+5)!}{(n+7)!} = \frac{1}{(n+6)(n+7)} = \frac{1}{n^2
+ 13n + 42}
\end{equation*}
The asymptotic order of the terms is \(\frac{1}{n^2}\text{,}\) which are the terms of a convergent \(\zeta\) series. Therefore, this series converges.
Activity 10.4.21.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{20 - n^2}{30 - n^3}
\end{equation*}
Activity 10.4.22.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{n^6 + 5n + 4}{7n^8 + 4n^3}
\end{equation*}
Activity 10.4.23.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{1}{3n - \sqrt{n}}
\end{equation*}
Activity 10.4.24.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^{\infty} \frac{n}{3^n}
\end{equation*}
Solution.
I will use the ratio test. I calculate the limit of the ratio of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{\left| \frac{n+1}{3^{n+1}}
\right|}{\left| \frac{n}{3^n} \right|} = \lim_{n \rightarrow
\infty} \frac{n+1}{n} \frac{3^n}{3^{n+1}} = \frac{1}{3}
\end{equation*}
The limit is less than one, so the series converges.
Activity 10.4.25.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \left( \frac{1}{n} + \frac{1}{n^2} +
\frac{1}{n^3} \right)
\end{equation*}
Solution.
Since this series is formed byadding positive elements to \(\frac{1}{n}\text{,}\) the terms here satisfy this inequatily.
\begin{equation*}
\left( \frac{1}{n} + \frac{1}{n^2} +
\frac{1}{n^3} \right) > \frac{1}{n}
\end{equation*}
The terms are positive and greater than the terms of the harmonic series, which is a divergent series with positive terms. By direct comparison, this series diverges.
Activity 10.4.26.
Test this series for convergence.
\begin{equation*}
\sum_{n=1}^\infty \frac{\sqrt{n^6 + 1}}{n^3}
\end{equation*}
Solution.
I can use the test for divergence here and look at the limit of the terms.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{\sqrt{n^6 + 1}}{n^3}
\end{equation*}
In this limit, the asymptotic order of the numerator and denominator are both \(n^3\) and the leading coefficients are both 1.
\begin{equation*}
\lim_{n \rightarrow \infty} \frac{\sqrt{n^6 + 1}}{n^3} =
\frac{1}{1} = 1
\end{equation*}
Since the limit of the terms is not zero, this is a divergent series.
Activity 10.4.27.
Test this series for convergence.
\begin{equation*}
\sum_{n=5}^\infty \frac{1}{\sqrt{3n^3-4}}
\end{equation*}
Solution.
The terms of this series are asymptotically equivalent to \(\frac{1}{n^{\frac{3}{2}}}\text{.}\) Those are the terms of a \(\zeta\) series with \(p=\frac{3}{2}\text{.}\) \(\zeta\) series converge when \(p > 1\text{,}\) so the series has terms which are asymptotically equivalent to a convergent \(\zeta\) series; therefore the series converges.
Activity 10.4.28.
Test this series for convergence.
\begin{equation*}
\sum_{n=3}^\infty \frac{1}{(\ln n)^n}
\end{equation*}
Solution.
This is a very strange series and it isn’t obvious what test might help. I found a comparison that works. For large enough \(n\text{,}\) \(\ln n \gt 2\text{.}\) If I invert this, I see that \(\frac{1}{\ln n} \lt \frac{1}{2}\text{.}\) Then if I apply the exponent \(n\text{,}\) I get
\begin{equation*}
\frac{1}{(\ln n)^n} \lt \left( \frac{1}{2} \right)^n
\end{equation*}
This shows that the terms of this series are smaller than the terms \(\left( \frac{1}{2} \right)^n\text{,}\) which are the terms of a convergent geometric series (since the common ratio, \(\frac{1}{2}\text{,}\) is less than 1). Since all the terms of our series and the comparison series are positive, since there are terms which are smaller than the terms of a convergent series, this series will also converge.
Alternatively, the root test is quite effective here (though I completely forgot about it while looking for the comparison above).
\begin{equation*}
\lim_{n \rightarrow \infty} \sqrt[n]{ \left| \frac{1}{(\ln
n)^n} \right| } = \lim_{n \rightarrow \infty} \frac{1}{\ln
n} = 0
\end{equation*}
Since this limits exists and is less than \(1\text{,}\) the root test tells me that the series converges.
