Lecture 36 Review 2

Exercise 36.1 Determine if the following series converge or diverge. In the case of a convergent series, find the sum of the series.

  1. \(\sum_{n=0}^{\infty} \frac{\cos{(\pi n)}}{(\sin{\left(\frac{\pi}{3}\right)})^n}\)
  2. \(\displaystyle \sum_{n=0}^{\infty} \frac{3^{n+1}+2^{2n}}{7^n}\)

  2. We can re-write the series as \[\sum_{n=0}^{\infty} \frac{\cos{(\pi n)}}{(\sin{\left(\frac{\pi}{3}\right)})^n}=\sum_{n=0}^{\infty}(-1)^n\left(\frac{2}{\sqrt{3}}\right)^n\] We get then that \(\displaystyle r= \frac{-2}{\sqrt{3}}\) and so, \(|r|>1\) which gives that this geometric series diverges.

  3. We can re-write the series as \[ \sum_{n=0}^{\infty} \frac{3^{n+1}+2^{2n}}{7^n}=\sum_{n=0}^{\infty}\left( 3\left(\frac{3}{7}\right)^n+\left(\frac{4}{7}\right)^n\right)\] We have that the series \[\sum_{n=0}^{\infty}3\left(\frac{3}{7}\right)^n\] converges since \(\displaystyle \frac{3}{7} <1\). The sum is \[\sum_{n=0}^{\infty}3\left(\frac{3}{7}\right)^n=\frac{3}{1-\frac{3}{7}}=\frac{21}{4}\] We also have the series \[\sum_{n=0}^{\infty}\left(\frac{4}{7}\right)^n\] converges since \(\displaystyle \frac{4}{7}<1\). The sum is \[\sum_{n=0}^{\infty}\left(\frac{4}{7}\right)^n=\frac{1}{1-\frac{4}{7}}=\frac{7}{3}\] From this we get that \(\displaystyle \sum_{n=0}^{\infty}\left( 3\left(\frac{3}{7}\right)^n+\left(\frac{4}{7}\right)^n\right)\) converges and the sum is \[\sum_{n=0}^{\infty} \frac{3^{n+1}+2^{2n}}{7^n}=\frac{21}{4}+\frac{7}{3}=\frac{91}{12}\]

Exercise 36.2 Test the following series for convergence by making a suitable comparison.

  1. \(\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+2n+2}\)
  2. \(\displaystyle \sum_{n=1}^{\infty} \sin\bigg(\frac{1}{n}\bigg)\)
  3. \(\displaystyle \sum_{n=1}^{\infty} \frac{3^{1/n}}{n^3}\)
  4. \(\displaystyle \sum_{n=1}^{\infty} \frac{\arctan n}{n}\)
  5. \(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}\)
  2. For positive \(n\), \(n^2 + 2n + 2 > n^2\), so:

\[ \frac{\sqrt{n}}{n^2 + 2n + 2} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}} \]

and \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}\) converges (p-series test), so by the comparison test \(\displaystyle \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+2n+2}\) also converges.

  1. For small values of the argument, \(\sin(z) \approx z\). Formally, \(\displaystyle \lim_{z \to 0} \frac{\sin(z)}{z} = 1\). Letting \(z = \frac{1}{n}\), \(z \to 0 = n \to \infty\), so

\[ \lim_{n \to \infty} \frac{\sin\left(\frac{1}{n}\right)}{ \frac{1}{n}} = 1 \]

and \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}\) diverges, so by the limit comparison test, \(\displaystyle \sum_{n=1}^{\infty} \sin \bigg(\frac{1}{n} \bigg)\) diverges.

  1. \(\displaystyle \lim_{n \to \infty} \frac{ \frac{3^{1/n}}{n^3}}{\frac{1}{n^3}} = \lim_{n \to \infty} 3^{1/n} = 1\), and \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3}\) converges, so by the limit comparison test \(\displaystyle \sum_{n=1}^{\infty} \frac{3^{1/n}}{n^3}\) converges

\(3^{1/n} < 3\), so \(\displaystyle \frac{3^{1/n}}{n^3} < \frac{3}{n^3}\), and \(\displaystyle \sum_{n=1}^{\infty} \frac{3}{n^3}\) converges, so by the comparison test} \(\displaystyle \sum_{n=1}^{\infty} \frac{3^{1/n}}{n^3}\) converges

  1. \(\displaystyle \lim_{n \to \infty} \frac{ \frac{\arctan n}{n}}{\frac{1}{n}} = \lim_{n \to \infty} \arctan{n} = \frac{\pi}{2}\), and \(\displaystyle \sum_{n=1}^{\infty}{1} \frac{1}{n}\) diverges, so by the limit comparison test \(\displaystyle \sum_{n=1}^{\infty} \frac{\arctan n}{n}\) diverges.

  2. \[ \frac{n!}{n^n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n-1}{n} \cdot \frac{n}{n} \]

Each of these terms is less than or equal to 1, so \(\dfrac{n!}{n^n} \leq \dfrac{1}{n} \cdot \dfrac{2}{n} = \dfrac{2}{n^2}\), provided \(n \geq 2\), so since \(\displaystyle \sum_{n=1}^{\infty}{2} \frac{2}{n^2}\) converges, by the comparison test \(\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}\) converges.

Exercise 36.3 Find the radius and interval of convergence of

  1. \(\displaystyle \sum_{n=1}^{\infty}\frac{8^n}{n}(2x+1)^n\).
  2. \(\displaystyle \sum_{n=0}^{\infty}\frac{n^2}{3^{2n}}(x+2)^n\)
  3. \(\displaystyle \sum_{n=0}^{\infty}\frac{n+1}{n!}(x-1)^n\).
  4. \(\displaystyle\sum_{n=1}^\infty (-1)^n \frac{(x-3)^n}{2n+1}\)
  5. \(\displaystyle \sum_{n=1}^\infty \frac{10^n x^n}{n^{10}}\)
  2. The ratio test gives \[\begin{align*} \lim_{n\to \infty}\Bigg| \frac{8^{n+1}(2x+1)^{n+1}}{n+1}\cdot\frac{n}{8^n(2x+1)^n}\Bigg|&=\lim_{n\to \infty}\Bigg|\frac{n}{n+1}8(2x+1)\Bigg|\\ &=8\Big|2x+1\Big| \end{align*}\] To converge we must have that \(8\big|2x+1\big|<1\). That is, we must require that \[\begin{align*} \big|2x+1\big|&<\frac{1}{8}\\ 2\Big|x+\frac{1}{2}\Big|&<\frac{1}{8}\\ \Big|x+\frac{1}{2}\Big|&<\frac{1}{16} \end{align*}\] We see then that the radius of convergence is \(\displaystyle \frac{1}{16}\). The centre of the series is \(\displaystyle x=-\frac{1}{2}\) so the endpoints are \(\displaystyle x=-\frac{9}{16}\) and \(\displaystyle x=-\frac{7}{16}\). We now check for convergence at the endpoints.

At \(\displaystyle x=-\frac{9}{16}\), we get \[\sum_{n=1}^{\infty}\frac{8^n}{n}\left(\frac{-1}{8}\right)^n=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\] which converges by the alternating series test.

At \(\displaystyle x=-\frac{7}{16}\) we get \[\sum_{n=1}^{\infty}\frac{8^n}{n}\left(\frac{1}{8}\right)^n=\sum_{n=1}^{\infty}\frac{1}{n}\] which diverges since it is the Harmonic series.

Thus, the interval of convergence is \(\displaystyle\left[-\frac{9}{16},-\frac{7}{16} \right)\).

  1. The ratio test gives \[\begin{align*} \lim_{n\to \infty}\Bigg| \frac{(n+1)^2(x+2)^{n+1}}{3^{2n+2}}\cdot\frac{3^{2n}}{n^2(x+2)^n}\Bigg|&=\lim_{n\to \infty}\Bigg|\frac{(n+1)^2}{9n^2}(x+2)\Bigg|\\ &=\frac{1}{9}|x+2| \end{align*}\] To converge we must have that \(\displaystyle \frac{1}{9}|x+2|<1\). That is, we must require that \(|x+2|<9\). We see that the radius of convergence is 9. The centre of the series is \(x=-2\) so the endpoints are \(x=-11\) and \(x=7\). We now check for convergence at the endpoints.

At \(x=-11\), we get \[ \sum_{n=0}^{\infty}\frac{n^2}{3^{2n}}(-9)^n=\sum_{n=0}^{\infty}(-1)^nn^2\] which diverges by the divergence test.

At \(x=7\), we get \[ \sum_{n=0}^{\infty}\frac{n^2}{3^{2n}}9^n= \sum_{n=0}^{\infty}n^2\] which diverges by the divergence test.

Thus, the interval of convergence is \((-11,7)\).

  1. The ratio test gives \[\begin{align*} \lim_{n\to \infty}\Bigg| \frac{(n+2)(x-1)^{n+1}}{(n+1)!}\cdot \frac{n!}{(n+1)(x-1)^n}\Bigg|&=\lim_{n\to\infty}\Bigg| \frac{n+2}{(n+1)^2}(x-1)\Bigg|\\ &=\lim_{n\to\infty}\Bigg| \frac{\frac{n}{n^2}+\frac{2}{n^2}}{\frac{n^2}{n^2}+\frac{2n}{n^2}+\frac{1}{n^2}}(x-1)\Bigg|\\ &=0 \end{align*}\] Thus, the radius of convergence is \(\infty\) and the interval of convergence is \(\mathbb{R}\).

  2. By the ratio test: \[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\cdot\frac{2n+1}{(-1)^n(x-3)^n}\right| = |x-3| \]

Now, this limit is less than 1 (and hence the series converges) whenever \(|x-3|<1\), and the limit is greater than 1 (and hence the series diverges) whenever \(|x-3|>1\). The radius of convergence is \(1\). Now test the endpoints \(x = 1+3=4\) and \(x = -1+3 = 2\).

If \(x = 4\), then the series becomes \(\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{(4-3)^n}{2n+1} = \sum_{n=1}^\infty (-1)^n \frac{1}{2n+1}}\), which converges by the alternating series test.

If \(x = 2\), then the series becomes \(\displaystyle{\sum_{n=1}^\infty (-1)^n \frac{(2-3)^n}{2n+1} = \sum_{n=1}^\infty (-1)^n \frac{(-1)^n}{2n+1}} = \displaystyle{ \sum_{n=1}^\infty \frac{1}{2n+1}}\), which diverges (this can be proven with a limit comparison with \(\frac{1}{n}\)).

Hence, the radius of convergence is 1, and the interval of convergence is \(x\in(2,4]\).

  1. By the Ratio Test: \[ \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{10^{n+1} x^{n+1}}{(n+1)^{10}}\cdot \frac{n^{10}}{10^n x^n}\right| = 10 \left(\frac{n}{n+1}\right)^{10} |x| = 10 \left(\frac{1}{1+1/n}\right)^{10} |x| = 10|x| \]

To converge, this must be less than 1, hence \(\displaystyle |x| < \frac{1}{10}\). Therefore the radius of convergence is \(\frac{1}{10}\). Now test the endpoints \(\pm \frac{1}{10}\).

At \(x=\frac{1}{10}\), the series is \(\displaystyle \sum_{n=1}^\infty \frac{10^n (1/10)^n}{n^{10}} = \sum_{n=1}^\infty \frac{1}{n^{10}}\) which is a convergent \(p\)-series.

At \(x=-\frac{1}{10}\), the series is \(\displaystyle \sum_{n=1}^\infty \frac{10^n (-1/10)^n}{n^{10}} = \sum_{n=1}^\infty \frac{(-1)^n}{n^{10}}\) which is convergent by the alternating series test.

The interval of convergence is therefore \(\displaystyle \left[-\frac{1}{10},\frac{1}{10}\right]\).

Exercise 36.4 Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

  1. \(\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}}\)
  2. \(\displaystyle{\sum_{n=1}^\infty \frac{n!}{e^n}}\)
  3. \(\displaystyle{\sum_{n=1}^\infty(-1)^{n} \frac{n}{\sqrt{n^3+2}}}\)
  4. \(\displaystyle{\sum_{n=1}^\infty(-1)^{n+1} \frac{n^2 \,2^n}{n!}}\)
  5. \(\displaystyle \sum_{n=1}^{\infty} \frac{\cos(1/n)}{n^2}\)

  2. We consider the series of absolute values, \(\displaystyle{\sum_{n=1}^\infty\frac{1}{n^4}}\), which clearly converges (\(p\)-series with \(p=4>1\)). Hence the series \(\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^4}}\) converges absolutely.

  3. We use the ratio test with \(a_n = \frac{n!}{e^n}\). Then

\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)!}{e^{n+1} }\cdot \frac{e^n}{n!} = \lim_{n\to\infty}\frac{n+1}{e} = \infty.\]

Therefore, by the ratio test, the series diverges.

  1. To check for absolute convergence, focus on the series of absolute values: \(\displaystyle{\sum_{n=1}^\infty \frac{n}{\sqrt{n^3+2}}}\). We can show this diverges using the limit comparison test, comparing it with \(\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{2}}}}\) (this is a \(p\)-series with \(p = 1/2 <1\), and diverges). That is,

\[ \lim_{n\to\infty} \frac{\left(\frac{n}{\sqrt{n^3+2}}\right)}{\left(\frac{1}{\sqrt n} \right)} = \lim_{n\to\infty}\frac{n\sqrt{n}}{\sqrt{n^3+2}} = \lim_{n\to\infty}\frac{\sqrt{n^3}}{\sqrt{n^3+2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1+2/n^3}} = 1 \]

Since the limit is a finite, non-zero number, we conclude that \(\displaystyle{\sum_{n=1}^\infty \frac{n}{\sqrt{n^3+2}}}\) diverges, so the original alternating series \(\displaystyle{\sum_{n=1}^\infty(-1)^{n} \frac{n}{\sqrt{n^3+2}}}\) does not absolutely converge. It remains to check if it converges conditionally.

Let \(b_n = \frac{n}{\sqrt{n^3+2}}\). We must check the conditions of the alternating series test. Clearly \(b_n\) is positive and \(\lim_{n\to\infty}b_n = 0\) (since the degree of the denominator is greater than that of the numerator).

To check if \(b_n\) is decreasing, consider the function \(f(x) = \frac{x}{\sqrt{x^3+2}}\). Then

\[ f^\prime(x) = \frac{\sqrt{x^3+2} - x\left(\frac{1}{2\sqrt{x^3+2}}\right)(3x^2)}{x^3+2} = \frac{2(x^3+2) - 3x^3}{2(x^3+2)^{3/2}} = \frac{4-x^3}{2(x^3+2)^{3/2}} \]

This derivative is negative for \(x>4^{1/3}\), meaning that the function is decreasing. We can now conclude that \(b_n\) is a decreasing sequence for all \(n\in\mathbb N\), \(n\ge 2\). By the alternating series test, \(\displaystyle{\sum_{n=1}^\infty(-1)^{n} \frac{n}{\sqrt{n^3+2}}}\) converges.

Therefore, the series \(\displaystyle{\sum_{n=1}^\infty(-1)^{n} \frac{n}{\sqrt{n^3+2}}}\) converges conditionally.

  1. We apply the ratio test. Here \(a_n = (-1)^{n+1}\frac{n^2\, 2^n}{n!}\), so

\[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{(-1)^{n+2}(n+1)^2\,2^{n+1}\,n!}{(-1)^{n+1}(n+1)!\,n^2\,2^n}\right| = \lim_{n\to\infty}\frac{(n+1)^2\,2}{(n+1)\,n^2} = 2\lim_{n\to\infty}\frac{(n+1)}{n^2} = 0. \]

Therefore, since this limit is less than 1, by the ratio test, the series converges absolutely.

  1. The series of absolute values is \(\displaystyle \sum_{n=1}^{\infty} \frac{|\cos(1/n)|}{n^2}\). Since \(|\cos(1/n)| \le 1\), then \(\frac{|\cos(1/n)|}{n^2} \le \frac{1}{n^2}\). Since \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}\) is a convergent \(p\)-series, then by the Comparison Test, \(\displaystyle \sum_{n=1}^{\infty} \frac{|\cos(1/n)|}{n^2}\) is convergent also.

Since the series of absolute values is convergent, then \(\displaystyle \sum_{n=1}^{\infty} \frac{\cos(1/n)}{n^2}\) is absolutely convergent.