Calculus at Moravian University

1. Since \(r=3/4\lt 1\text{,}\) this series converges. By Theorem 9.2.5, we have that

   \begin{equation*} \infser[0] \left(\frac34\right)^n = \frac{1}{1-3/4} = 4. \end{equation*}

   However, note the subscript of the summation in the given series: we are to start with \(n=2\text{.}\) Therefore we subtract off the first two terms, giving:

   \begin{equation*} \sum_{n=2}^\infty \left(\frac34\right)^n = 4 - 1 - \frac34 = \frac94. \end{equation*}

   This is illustrated in Figure 9.2.8.(a).

   

2. Since \(\abs{r} = 1/2 \lt 1\text{,}\) this series converges, and by Theorem 9.2.5,

   \begin{equation*} \infser[0] \left(\frac{-1}{2}\right)^n = \frac{1}{1-(-1/2)} = \frac23. \end{equation*}

   The partial sums of this series are plotted in Figure 9.2.8.(a). Note how the partial sums are not purely increasing as some of the terms of the sequence \(\{(-1/2)^n\}\) are negative.

3. Since \(r \gt 1\text{,}\) the series diverges. (This makes “common sense”; we expect the sum

   \begin{equation*} 1+3+9+27 + 81+243+\cdots \end{equation*}

   to diverge.) This is illustrated in Figure 9.2.8.(b).

   

   

1. This is a \(p\)–series with \(p=1\text{.}\) By Theorem 9.2.10, this series diverges. This series is a famous series, called the Harmonic Series, so named because of its relationship to harmonics in the study of music and sound.

2. This is a \(p\)–series with \(p=2\text{.}\) By Theorem 9.2.10, it converges. Note that the theorem does not give a formula by which we can determine what the series converges to; we just know it converges. A famous, unexpected result is that this series converges to \(\ds{\pi^2}/{6}\text{.}\)

3. This is a \(p\)–series with \(p=1/2\text{;}\) the theorem states that it diverges.

4. This is not a \(p\)–series; the definition does not allow for alternating signs. Therefore we cannot apply Theorem 9.2.10. (Another famous result states that this series, the Alternating Harmonic Series, converges to \(\ln(2)\text{.}\))

5. This is a general \(p\)–series with \(p=3\text{,}\) therefore it converges.

6. This is not a \(p\)–series, but a geometric series with \(r=1/2\text{.}\) It converges.

It will help to write down some of the first few partial sums of this series.

Note how most of the terms in each partial sum are canceled out! In general, we see that \(\ds S_n = 1-\frac{1}{n+1}\text{.}\) The sequence \(\{S_n\}\) converges, as \(\lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\left(1-\frac1{n+1}\right) = 1\text{,}\) and so we conclude that \(\ds \infser \left(\frac1n-\frac1{n+1}\right) = 1\text{.}\) Partial sums of the series are plotted in Figure 9.2.13.

1. We can decompose the fraction \(2/(n^2+2n)\) as

   \begin{equation*} \frac2{n^2+2n} = \frac1n-\frac1{n+2}. \end{equation*}

   (See Section 6.5, Partial Fraction Decomposition, to recall how this is done, if necessary.) Expressing the terms of \(\{S_n\}\) is now more instructive:

   \begin{align*} S_1 \amp = 1-\frac13 \amp \amp = 1-\frac13\\ S_2 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right) \amp \amp = 1+\frac12-\frac13-\frac14\\ S_3 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right) \amp \amp = 1+\frac12-\frac14-\frac15\\ S_4 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right) \amp \amp = 1+\frac12-\frac15-\frac16\\ S_5 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right) \amp \amp = 1+\frac12-\frac16-\frac17 \end{align*}

   We again have a telescoping series. In each partial sum, most of the terms cancel and we obtain the formula \(\ds S_n = 1+\frac12-\frac1{n+1}-\frac1{n+2}\text{.}\) Taking limits allows us to determine the convergence of the series: \(\lim_{n\to\infty}S_n = \lim_{n\to\infty} \left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right) = \frac32\) So \(\infser \frac1{n^2+2n} = \frac32\text{.}\) This is illustrated in Figure 9.2.15.(a).

   

   

2. We begin by writing the first few partial sums of the series:

   \begin{align*} S_1 \amp = \ln\left(2\right)\\ S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right)\\ S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)\\ S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) \end{align*}

   At first, this does not seem helpful, but recall the logarithmic identity: \(\ln(x) +\ln(y) = \ln(xy)\text{.}\) Applying this to \(S_4\) gives:

   \begin{align*} S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)\\ \amp= \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right)\\ \amp = \ln\left(5\right). \end{align*}

   We can conclude that \(\{S_n\} = \big\{\ln(n+1)\big\}\text{.}\) This sequence does not converge, as \(\lim\limits_{n\to\infty}S_n=\infty\text{.}\) Therefore \(\ds\infser \ln\left(\frac{n+1}{n}\right)=\infty\text{;}\) the series diverges. Note in Figure 9.2.15(b) how the sequence of partial sums grows slowly; after 100 terms, it is not yet over 5. Graphically we may be fooled into thinking the series converges, but our analysis above shows that it does not.

1. We start by using algebra to break the series apart:

   \begin{align*} \infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} \amp = \infser\left(\frac{(-1)^{n+1}n^2}{n^3}-\frac{(-1)^{n+1}n}{n^3}\right)\\ \amp = \infser\frac{(-1)^{n+1}}{n}-\infser\frac{(-1)^{n+1}}{n^2}\\ \amp = \ln(2) - \frac{\pi^2}{12} \approx -0.1293. \end{align*}

   This is illustrated in Figure 9.2.19.(a).

2. This looks very similar to the series that involves \(e\) in Key Idea 9.2.17. Note, however, that the series given in this example starts with \(n=1\) and not \(n=0\text{.}\) The first term of the series in the Key Idea is \(1/0! = 1\text{,}\) so we will subtract this from our result below:

   \begin{align*} \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!}\\ \amp = 1000\cdot (e-1) \approx 1718.28. \end{align*}

   This is illustrated in Figure 9.2.19.(b). The graph shows how this particular series converges very rapidly.

   

   

3. The denominators in each term are perfect squares; we are adding \(\ds \sum_{n=4}^\infty \frac{1}{n^2}\) (note we start with \(n=4\text{,}\) not \(n=1\)). This series will converge. Using the formula from Key Idea 9.2.17, we have the following:

   \begin{align*} \infser \frac1{n^2} \amp = \sum_{n=1}^3 \frac1{n^2} +\sum_{n=4}^\infty \frac1{n^2}\\ \infser \frac1{n^2} - \sum_{n=1}^3 \frac1{n^2} \amp =\sum_{n=4}^\infty \frac1{n^2}\\ \frac{\pi^2}{6} - \left(\frac11+\frac14+\frac19\right) \amp = \sum_{n=4}^\infty \frac1{n^2}\\ \frac{\pi^2}{6} - \frac{49}{36} \amp = \sum_{n=4}^\infty \frac1{n^2}\\ 0.2838\amp \approx \sum_{n=4}^\infty \frac1{n^2} \end{align*}