OBJECTIVES

When you finish this section, you should be able to:

1. Use Comparison Tests for Convergence and Divergence (p. 576)
2. Use the Limit Comparison Test (p. 577)

RECALL

The General Convergence Test: An infinite series of positive terms converges if and only if its sequence of partial sums is bounded.

THEOREM Comparison Test for Convergence

If $0<a_{k}\leq b_{k}$ for all $k$ and $\sum\limits_{k=1}^{\infty}b_{k}$ converges, then $\sum\limits_{k=1}^{\infty}a_{k}$ converges.

Show that the series below converges: $$\sum\limits_{k=1}^{\infty }\frac{1}{k^{k}}=1+\frac{1}{2^{2}}+\frac{1}{3^{3}} +\cdots +\frac{1}{n^{n}}+\cdots$$

Solution We know that the geometric series $\sum\limits_{k=1}^{\infty }\dfrac{1}{2^{k}}$ converges since $\vert r\vert =\dfrac{1}{2}<1$. Now, since $\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}$ for all $n\geq 2$, except for the first term, each term of the series $\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}$ is less than or equal to the corresponding term of the convergent geometric series. So, by the Comparison Test for Convergence, the series $\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}$ converges.

NOTE

Do you see why $\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}$ when $n \geq 2$? $n\geq 2\Rightarrow {n}^{n}\geq {2}^{n}\Rightarrow$ $\dfrac{{ 1}}{{n}^{n}}\leq \dfrac{1}{{2}^{n}}$.

NOW WORK

Problem 5.

THEOREM Comparison Test for Divergence

If $0<c_{k}\leq a_{k}$ for all $k$ and $\sum\limits_{k=1}^{\infty}c_{k}$ diverges, then $\sum\limits_{k=1}^{\infty }a_{k}$ diverges.

Show that the series $\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2)}$ diverges.

Solution Since $\dfrac{k+3}{k(k+2)}$ has a factor $\dfrac{1}{k}$, we choose to compare the given series to the harmonic series $\sum\limits_{k=1}^{\infty }\dfrac{1}{k}$, which diverges. $$\frac{n+3}{n(n+2)}=\left( {\frac{n+3}{n+2}}\right) \left( {\frac{1}{n}} \right) >\frac{1}{n}$$

It follows from the Comparison Test for Divergence that $\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2) }$ diverges.

NOW WORK

Problem 9.

THEOREM Limit Comparison Test

Suppose $\sum\limits_{k=1}^{\infty }\,a_{k}$ and $\sum\limits_{k=1}^{\infty }\,b_{k}$ are both series of positive terms.

If $\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L$, $0<$ $L<\infty ,$ then both series converge or both series diverge.

The Limit Comparison Test gives no guidance if $\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}$ $=0$ or $\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}=\infty$ or $\lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}$ does not exist.

Proof

If $\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L\,{>}\,0$, then for any $\varepsilon\,{>}\,0,$ there is a positive real number $N$ so that $\left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert <\varepsilon ,$ for all $n>N.$ If we choose $\varepsilon =\dfrac{L}{2},$ then $$\begin{eqnarray*} \begin{array}{rl@{\qquad\hspace{18.2pt}}l} \left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert &< \dfrac{L}{2} & \hbox{for all }n>N \\[13pt] \dfrac{L}{2}<\dfrac{a_{n}}{b_{n}} &< \dfrac{3L}{2} & \hbox{for all }n>N\, \end{array} \end{eqnarray*}$$

Since $b_{n}>0$, $$\begin{array}{rl@{\qquad}l} \dfrac{L}{2}b_{n}<a_{n} & <\dfrac{3L}{2}b_{n} & \hbox{for all }n>N \end{array}$$

If $\sum\limits_{k=1}^{\infty }a_{k}$ converges, by the Comparison Test for Convergence, the series $\sum\limits_{k=1}^{\infty }\left(\dfrac{L}{2}b_{k}\right)$ $=\dfrac{L}{2}\sum\limits_{k=1}^{\infty }b_{k}$ also converges. It follows that the series $\sum\limits_{k=1}^{\infty }b_{k}$ is also convergent.

If $\sum\limits_{k=1}^{\infty }b_{k}$ converges, then so does $\sum\limits_{k=1}^{\infty }\left(\dfrac{3L}{2}b_{k}\right).$ Since $a_{n}<\dfrac{3L}{2}b_{n},$ for all $n>N,$ by the Comparison Test for Convergence, $\sum\limits_{k=1}^{\infty }a_{k}$ also converges.

Therefore, $\sum\limits_{k=1}^{\infty }a_{k}$ and $\sum\limits_{k=1}^{\infty}b_{k}$ converge together. Since one cannot converge and the other diverge, they must diverge together.

RECALL

If $c$ is a nonzero real number and if $\sum\limits_{k\,=\,1}^{\infty}{a}_{k}={S}$ is a convergent series, then the series $\sum\limits_{k\,=\,1}^{\infty} ({ca}_{k})$ also converges. Furthermore, $\sum\limits_{k\,=\,1}^{\infty}({ca}_{k})={c}\sum\limits_{k\,=\,1}^{\infty }{a}_{k}={cS}$.

Determine whether the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}$ converges or diverges.

Solution We choose an appropriate $p$-series to use for comparison by examining the behavior of the series for large values of $n$: $$\begin{eqnarray*} \dfrac{1}{2n^{3/2}+5}=\dfrac{1}{n^{3/2}\left( 2+\dfrac{5}{n^{3/2}}\right) }= \dfrac{1}{n^{3/2}}\left( \dfrac{1}{2+\dfrac{5}{n^{3/2}}}\right) & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow }}}} }{\approx} & \dfrac{1}{n^{3/2}}\left(\dfrac{1}{2}\right)\\ \end{eqnarray*}$$

This leads us to choose the $p$-series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}},$ which converges, and use the Limit Comparison Test with $$a_{n}=\dfrac{1}{2n^{3/2}+5} \qquad \hbox{and}\qquad b_{n}=\dfrac{1}{n^{3/2}}$$ $$\lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}=\lim_{n\,\rightarrow \,\infty }\frac{\dfrac{1}{2n^{3/2}+5}}{\dfrac{1}{n^{3/2}}} =\lim\limits_{n\rightarrow \infty }\frac{n^{3/2}}{2n^{3/2}+5} =\lim\limits_{n\rightarrow \infty }\dfrac{1}{2+\dfrac{5}{n^{3/2}}}=\frac{1}{2}$$

Since the limit is a positive number and the $p$-series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}}$ converges, then by the Limit Comparison Test, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}$ also converges.

NOW WORK

Problem 15.

Determine whether the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k}+2}{\sqrt{k^{3}+3k^{2}+1}}$ converges or diverges.

Solution We choose a $p$-series for comparison by examining how the terms of the series behave for large values of $n$: $$\begin{eqnarray*} \frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}=\frac{\sqrt{n}\,\left( 3+\dfrac{2}{ \sqrt{n}}\right) }{\sqrt{n^{3}}\left( \sqrt{1+\dfrac{3}{n}+\dfrac{1}{n^{3}}} \right) }=\dfrac{n^{1/2}}{n^{3/2}}\frac{3+\dfrac{2}{\sqrt{n}}}{\sqrt{1+ \dfrac{3}{n}+\dfrac{1}{n^{3}}}} & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow}}}}}{\approx} & \dfrac{1}{n}(3) \\ \end{eqnarray*}$$

So, we compare the series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}$ to the harmonic series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}$, which diverges, and use the Limit Comparison Test with $$a_{n}=\frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}\qquad \hbox{ and }\qquad b_{n}=\dfrac{1}{n}$$ $$\begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}} &=& \lim_{n\,\rightarrow \,\infty }\frac{\dfrac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}}{\dfrac{1}{n}} =\lim\limits_{n\rightarrow \infty }\frac{n\left( 3\sqrt{n}+2\right) }{\sqrt{ n^{3}+3n^{2}+1}}=\lim\limits_{n\rightarrow \infty }\frac{3n^{3/2}+2n}{\sqrt{n^{3}+3n^{2}+1}}\\[7pt] &=&\lim\limits_{n\rightarrow \infty }\dfrac{3n^{3/2}}{n^{3/2}}=3 \end{eqnarray*}$$

Since the limit is a positive real number and the $p$-series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}$ diverges, then by the Limit Comparison Test, $\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}$ also diverges.

NOW WORK

Problem 17.