Active Calculus

1. The completed table is shown below.

   \(\sum_{k=1}^{1} (-1)^{k+1}\frac{1}{k}\)	\(1\)	\(\sum_{k=1}^{6} (-1)^{k+1}\frac{1}{k}\)	\(0.6166666667\)
   \(\sum_{k=1}^{2} (-1)^{k+1}\frac{1}{k}\)	\(0.5\)	\(\sum_{k=1}^{7} (-1)^{k+1}\frac{1}{k}\)	\(0.7595238095\)
   \(\sum_{k=1}^{3} (-1)^{k+1}\frac{1}{k}\)	\(0.8333333333\)	\(\sum_{k=1}^{8} (-1)^{k+1}\frac{1}{k}\)	\(0.6345238095\)
   \(\sum_{k=1}^{4} (-1)^{k+1}\frac{1}{k}\)	\(0.5833333333\)	\(\sum_{k=1}^{9} (-1)^{k+1}\frac{1}{k}\)	\(0.7456349206\)
   \(\sum_{k=1}^{5} (-1)^{k+1}\frac{1}{k}\)	\(0.7833333333\)	\(\sum_{k=1}^{10} (-1)^{k+1}\frac{1}{k}\)	\(0.6456349206\)

2. The entries in the sequence of partial sums from part (a) is shown in the figure below.

   

   Notice that the partial sums seem to oscillate back and forth around some fixed number, getting closer to that fixed number at each successive step. So there appears to be a limit for this sequence of partial sums.

1. Since \((k+1)^2 + 2 \gt k^2+2\) it follows that \(\frac{1}{(k+1)^2+2} \lt \frac{1}{k^2+2}\) and the sequence \(\frac{1}{k^2+2}\) decreases to 0. The Alternating Series Test shows then that the series \(\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}\) converges.

2. In this case we have that \(\lim_{k \to \infty} \frac{2k}{k+5} = 2\) and so the sequence of \(k\)th terms does not even converge to 0. So the series \(\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}k}{k+5}\) diverges by the Divergence Test.

3. We know that the natural log is an increasing function, so \(\frac{1}{\ln(k)}\) is a decreasing function. It is also true that \(\lim_{k \to \infty} \frac{1}{\ln(k)} = 0\) and so the series \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}\) converges by the Alternating Series Test.

First note that the sequence \(\left\{\frac{1}{k^4}\right\}\) decreases to 0, so the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4}\) converges by the Alternating Series Test. Let \(S\) be the sum \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4}\) and \(S_n\) the \(n\)th partial sum of the series. We know that

So we only need to determine the value of \(n\) so that

This happens when \((n+1)^4 \lt 10000\) or when \(n+1 \gt 10\text{.}\) So \(n = 10\) will do. A computer algebra system gives \(S_{10} \approx 0.9469925924\) while \(\frac{7}{720} \pi^4 \approx 0.9470328299\text{.}\) These do agree to \(0.0001\text{.}\)

1. Each term in the series

   \begin{equation*} 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \end{equation*}

   is of the form \(\frac{1}{k^2}\) or \(-\frac{1}{k^2}\text{.}\) Since each of these terms is less than or equal to \(\frac{1}{k^2}\) it follows that

   \begin{equation*} 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \lt \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} \end{equation*}

2. Each term in the series

   \begin{equation*} 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \end{equation*}

   is of the form \(\frac{1}{k^2}\) or \(-\frac{1}{k^2}\text{.}\) Since each of these terms is greater than or equal to \(-\frac{1}{k^2}\) it follows that

   \begin{equation*} 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \gt \sum_{k=1}^{\infty} -\frac{1}{k^2}\text{.} \end{equation*}

3. We know that \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is a \(p\)-series with \(p = 2 \gt 1\) and so is a convergent series. Since \(\sum_{k=1}^{\infty} \frac{1}{k^2} = -\sum_{k=1}^{\infty} \frac{1}{k^2}\text{,}\) the series \(\sum_{k=1}^{\infty} -\frac{1}{k^2}\) also converges. Since the terms in the series

   \begin{equation*} 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \end{equation*}

   converge to 0 and the series itself is bounded between two convergent series, we should expect this series to converge to some finite number between \(\sum_{k=1}^{\infty} -\frac{1}{k^2}\) and \(\sum_{k=1}^{\infty} \frac{1}{k^2}\text{.}\)

1. 1. By L'Hôpital's Rule we have

      \begin{equation*} \lim_{k \to \infty} \frac{\ln(k)}{k} = \lim_{k \to \infty} \frac{1}{k} = 0\text{.} \end{equation*}

      Also, \(\frac{d}{dk} \frac{\ln(k)}{k} = \frac{1-ln(k)}{k^2}\) is negative when \(k \gt e\text{,}\) so the sequence \(\left\{ \frac{\ln(k)}{k} \right\}\) ultimately decreases to 0. Since the first few terms in a series are irrelevant to its convergence or divergence, we conclude that the series \(\sum (-1)^k \frac{\ln(k)}{k}\) converges by the Alternating Series test.

   2. Note that

      \begin{align*} \lim_{t \to \infty} \int_{1}^{t} \frac{\ln(x}{x} \amp = \lim_{t \to \infty} \left. \frac{\ln(x)^2}{2} \right|_1^t\\ \amp = \lim_{t \to \infty} \frac{\ln(t)^2}{2}\\ \amp = \lim_{t \to \infty} \frac{\ln(t)^2}{2}\\ \amp = \infty\text{.} \end{align*}

      Since the improper integral diverges, the Integral Test shows that the series \(\sum (-1)^k \frac{\ln(k)}{k}\) diverges. So the series \(\sum (-1)^k \frac{\ln(k)}{k}\) converges conditionally.

   1. By L'Hôpital's Rule we have

      \begin{equation*} \lim_{k \to \infty} \frac{\ln(k)}{k^2} = \lim_{k \to \infty} \frac{1}{2k^2} = 0\text{.} \end{equation*}

      Also,

      \begin{equation*} \frac{d}{dk} \frac{\ln(k)}{k^2} = \frac{k-2k\ln(k)}{k^4} = \frac{1-2\ln(k)}{k^3} \end{equation*}

      is negative when \(k \gt e\text{,}\) so the sequence \(\left\{ \frac{\ln(k)}{k^2} \right\}\) ultimately decreases to 0. Since the first few terms in a series are irrelevant to its convergence or divergence, we conclude that the series \(\sum (-1)^k \frac{\ln(k)}{k^2}\) converges by the Alternating Series test.

   2. Notice that

      \begin{equation*} \lim_{k \to \infty} \frac{ \ln(k) }{ k^{1/2} } = \lim_{k \to \infty} \frac{2}{k^{1/2}} = 0\text{,} \end{equation*}

      So \(\frac{1}{\sqrt{k}}\) dominates \(\ln(k)\) and \(\ln(k) \lt \sqrt{k}\) for large \(k\text{.}\) It follows that

      \begin{equation*} \frac{\ln(k)}{k^2} \lt \frac{ \sqrt{k} }{k^2} = \frac{1}{k^{3/2}} \end{equation*}

      for large \(k\text{.}\) Therefore,

      \begin{equation*} \sum \frac{\ln(k)}{k^2} \lt \sum \frac{1}{k^{3/2}} \end{equation*}

      for large \(k\text{.}\) Since \(\sum \frac{1}{k^{3/2}}\) is a \(p\)-series with \(p=\frac{3}{2} \gt 1\text{,}\) the series \(\sum \frac{1}{k^{3/2}}\) converges. This forces the series \(\sum \frac{\ln(k)}{k^2}\) to converge as well. So \(\sum (-1)^k \frac{\ln(k)}{k^2}\) converges absolutely.

1. For large values of \(k\text{,}\) \(\frac{2}{\sqrt{k-2}}\) looks like \(\frac{2}{\sqrt{k}}\text{.}\) We will use the Limit Comparison Test to compare these two series:

   \begin{align*} \lim_{k \to \infty} \frac{\frac{2}{\sqrt{k-2}}}{\frac{2}{\sqrt{k}}} \amp = \lim_{k \to \infty} \frac{\sqrt{k}}{\sqrt{k-2}}\\ \amp = \lim_{k \to \infty} \sqrt{\frac{1}{1-\frac{2}{k}}}\\ \amp = 1\text{.} \end{align*}

   Since this limit is a positive constant, we know that the two series \(\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}\) and \(\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k}}\) either both converge or both diverge. Now \(\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k}} = 2\sum_{k=3}^{\infty} \ \frac{1}{k^{1/2}}\) is constant times a \(p\)-series with \(p=\frac{1}{2}\text{,}\) so \(\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k}}\) diverges. Therefore, \(\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}\) diverges as well.

2. Note that

   \begin{equation*} \lim_{k \to \infty} \frac{k}{1+2k} = \lim_{k \to \infty} \frac{k}{2k} = \frac{1}{2}\text{,} \end{equation*}

   so the Divergence Test shows that \(\sum_{k=1}^{\infty} \ \frac{k}{1+2k}\) diverges.

3. For large values of \(k\text{,}\) \(\frac{2k^2+1}{k^3+k+1}\) looks like \(\frac{2}{k}\text{.}\) We will use the Limit Comparison Test to compare these two series:

   \begin{align*} \lim_{k \to \infty} \frac{\frac{2k^2+1}{k^3+k+1}}{\frac{2}{k}} \amp = \lim_{k \to \infty} \frac{(2k^2+1)k}{2(k^3+k+1)}\\ \amp = \lim_{k \to \infty} \frac{2 + \frac{1}{k^2}}{2+\frac{1}{k^2}+\frac{1}{k^3}}\\ \amp = 1\text{.} \end{align*}

   Since this limit is a positive constant, we know that the two series \(\sum_{k=1}^{\infty} \ \frac{2k^2+1}{k^3+k+1}\) and \(\sum_{k=1}^{\infty} \ \frac{2}{k}\) either both converge or both diverge. Now \(\sum_{k=1}^{\infty} \ \frac{2}{k} = 2\sum_{k=1}^{\infty} \ \frac{1}{k}\) is constant times a \(p\)-series with \(p=1\text{,}\) so \(\sum_{k=1}^{\infty} \ \frac{2}{k}\) diverges. Therefore, \(\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}\) diverges as well.

4. Series that involve factorials are good candidates for the Ratio Test:

   \begin{align*} \lim_{k \to \infty} \frac{\frac{100^{k+1}}{(k+1)!}}{\frac{100^k}{k!}} \amp = \lim_{k \to \infty} \frac{100^{k+1}k!}{100^k(k+1)!}\\ \amp = \lim_{k \to \infty} \frac{100}{k+1}\\ \amp = 0\text{.} \end{align*}

   Since this limit is 0, the Ratio Test tells us that the series \(\sum_{k=0}^{\infty} \ \frac{100^k}{k!}\) converges.

5. Notice that \(\sum_{k=1}^{\infty} \ \frac{2^k}{5^k} = \sum_{k=1}^{\infty} \ \left(\frac{2}{5}\right)^k\) is a geometric series with ratio \(\frac{2}{5}\text{.}\) Let's write out the first few terms to identify the value of \(a\text{:}\)

   \begin{equation*} \sum_{k=1}^{\infty} \ \left(\frac{2}{5}\right)^k = \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots = \left(\frac{2}{5}\right)\left[ 1 + \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \cdots \right]\text{.} \end{equation*}

   So \(a = \frac{2}{5}\text{.}\) Since the ratio of this geometric series is between \(-1\) and 1, the series converges. The sum of the series is

   \begin{equation*} \sum_{k=1}^{\infty} \ \frac{2^k}{5^k} = \left(\frac{2}{5}\right) \left(\frac{1}{1-\frac{2}{5}} \right) = \left(\frac{2}{5}\right)\left(\frac{5}{3}\right) = \frac{2}{3}\text{.} \end{equation*}

6. For large values of \(n\text{,}\) \(\frac{k^3-1}{k^5+1}\) looks like \(\frac{k^3}{k^5}=\frac{1}{k^2}\text{.}\) We will use the Limit Comparison Test to compare \(\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}\) and \(\sum_{k=1}^{\infty} \ \frac{1}{k^2}\text{:}\)

   \begin{align*} \lim_{k \to \infty} \frac{\frac{k^3-1}{k^5+1}}{\frac{1}{k^2}} \amp = \lim_{k \to \infty} \frac{k^5-k^2}{k^5+1}\\ \amp = \lim_{k \to \infty} \frac{1-\frac{1}{k^3}}{1+\frac{1}{k^5}}\\ \amp = 1\text{.} \end{align*}

   Since this limit is 1, we know that the two series \(\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}\) and \(\sum_{k=1}^{\infty} \ \frac{1}{k^2}\) either both converge or both diverge. Now \(\sum_{k=1}^{\infty} \ \frac{1}{k^2}\) is a \(p\)-series with \(p=2\text{,}\) so \(\sum_{k=1}^{\infty} \ \frac{1}{k^2}\) converges. Therefore, \(\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}\) converges as well.

7. This series looks geometric, but to be sure let's write out the first few terms of this series:

   \begin{equation*} \sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k} = \frac{3}{7^2} + \frac{3^2}{7^3} + \frac{3^3}{7^4} + \cdots = \left( \frac{3}{7^2} \right) \left[1 + \frac{3}{7} + \left( \frac{3}{7} \right)^2 + \left( \frac{3}{7} \right)^3 + \cdots \right]\text{.} \end{equation*}

   So \(\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k}\) is a geometric series with \(a = \frac{3}{49}\) and \(r = \frac{3}{7}\text{.}\) Since the ratio of this geometric series is between \(-1\) and 1, the series converges. The sum of the series is

   \begin{align*} \sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k} \amp = \left(\frac{3}{49}\right) \left(\frac{1}{1-\frac{3}{7}} \right)\\ \amp = \left(\frac{3}{49}\right)\left(\frac{7}{4}\right)\\ \amp = \frac{3}{28}\text{.} \end{align*}

8. \(\sum_{k=2}^{\infty} \ \frac{1}{k^k}\) converges by comparison with the series \(\sum_{k=2}^{\infty} \ \frac{1}{k^2}\text{.}\)

9. This series is an alternating series of the form \(\sum_{k=1}^{\infty} (-1)^{k+1} a_k\) with \(a_k = \frac{1}{\sqrt{k+1}}\text{.}\) Since

   \begin{equation*} \lim_{k \to \infty} \frac{1}{\sqrt{k+1}} = 0 \end{equation*}

   and \(\frac{1}{\sqrt{k+1}}\) decreases to 0, the Alternating Series Test shows that \(\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}}\) converges.

10. For this example we will use the Integral Test with the substitution \(w = \ln(x)\text{,}\) \(dw = \frac{1}{x} \ dx\text{:}\)

    \begin{align*} \displaystyle \int_2^{\infty} \frac{1}{x \ln(x)} \ dx \amp = \lim_{b \to \infty} \int_2^b \frac{1}{x \ln(x)} \ dx\\ \amp = \displaystyle \lim_{b \to \infty} \int_{\ln(2)}^{\ln(b)} \frac{1}{w} \ dw\\ \amp = \displaystyle \lim_{b \to \infty} \ln(w) \bigm|_{\ln(2)}^{\ln(b)}\\ \amp = \displaystyle \lim_{b \to \infty} \left[ \ln(\ln(b)) - \ln(\ln(2)) \right]\\ \amp = \infty\text{.} \end{align*}

    Since \(\int_2^{\infty} \frac{1}{x \ln(x)} \ dx\) diverges, we conclude \(\sum_{k=2}^{\infty} \ \frac{1}{k \ln(k)}\) diverges.

11. First note that the terms \(\frac{1}{\ln(k)}\) decrease to \(0\text{,}\) so the Alternating Series Test shows that the alternating series \(\sum_{k=2}^{\infty} \frac{(-1)^k}{\ln(k)}\) converges. Let \(S = \sum_{k=2}^{\infty} \frac{(-1)^k}{\ln(k)}\) and let \(S_n\) be the \(n\)th partial sum of this series. Recall that

    \begin{equation*} |S_n - S| \lt |S_n - S_{n+1}| = a_{n+1} \end{equation*}

    where \(a_{n+1} = \frac{1}{\ln(n+1)}\text{.}\) If we can find a value of \(n\) so that \(a_{n+1} \lt 0.001\text{,}\) then \(|S_n - S| \lt 0.001\) as desired. Now

    \begin{align*} a_{n+1} \amp \lt 0.001\\ \frac{1}{\ln(n+1)} \amp \lt 0.001\\ \ln(n+1) \amp \gt 1000\\ n+1 \amp \gt e^{1000}\\ n \amp gt e^{1000}-1\text{.} \end{align*}

    So if we choose \(n = e^{1000}\text{,}\) then \(S_n\) approximates the sum of the series to within 0.001. Notice that \(e^{1000} \approx 1.97 \times 10^{434}\) is a very large number. This shows that the series \(\sum_{k=2}^{\infty} \frac{(-1)^k}{\ln(k)}\) converges very slowly.