11.2 Convergence Tests

In this section we prove a number of theorems about convergence of series of real numbers. Later we will show how to use these results to study convergence of complex sequences.

11.11   Theorem (Comparison test for series.) Let $f,g$ be two sequences of non-negative numbers. Suppose that there is a number $N\in\mbox{{\bf N}}$ such that

$$f(n)\leq g(n) \mbox{ for all }n\in\mbox{{\bf Z}}_{\geq N}.$$

Then

$$\mbox{ if } g\mbox{ is summable, then } f\mbox{ is summable, }$$

and

$$\mbox{ if } f\mbox{ is not summable, then } g\mbox{ is not summable. }$$

Proof: Note that the two statements in the conclusion are equivalent, so it is sufficient to prove the first.

Suppose that $g$ is summable. Then $\sum g$ converges, so $\sum g$ is bounded -- say $(\sum g)(n)\leq B$ for all $n\in\mbox{{\bf N}}$. Then for all $n\geq N+1$,

$$\begin{eqnarray*} \sum_{j=0}^nf(j)&=&\sum_{j=0}^Nf(j)+\sum_{j=N+1}^nf(j)\leq\sum... ...& \sum_{j=0}^N f(j) + \sum_{j=0}^n g(j) \leq \sum_{j=0}^Nf(j)+B. \end{eqnarray*}$$

Since for $n\leq N$ we have

$$\sum_{j=0}^nf(j)\leq\sum_{j=0}^Nf(j)\leq\sum_{j=0}^Nf(j)+B,$$

we see that $\sum f$ is bounded by $${\sum_{j=0}^Nf(j)+B}$$. Also $\sum f$ is increasing, since $(\sum f)(n+1)=(\sum f)(n)+f(n+1)\geq\sum f(n)$. Hence $\sum f$ is bounded and increasing, and hence $\sum f$ converges; i.e., $f$ is summable. $\mid\!\mid\!\mid$

11.12   Examples. Since

$${1\over {\sqrt n}}\geq{1\over n}>0\mbox{ for all } n\in\mbox{{\bf Z}}_{\geq 1}$$

and $${\sum\left\{{1\over n}\right\}_{n\geq 1}}$$ diverges, it follows that $${\sum\left\{{1\over {\sqrt n}}\right\}_{n\geq 1}}$$ also diverges. Since $\sum\{t^n\}$ converges for $0\leq t < 1$, $${\sum\left\{{{t^n}\over {n!}}\right\}}$$ also converges for $0\leq t < 1$.

In order to use the comparison test, we need to have some standard series to compare other series with. The next theorem will provide a large family of standard series.

11.13   Theorem. Let $p\in\mbox{{\bf Q}}$. Then $${ \left\{{1\over {n^p}}\right\}_{n\geq 1}}$$ is summable if $p>1$, and is not summable if $p\leq 1$.

Proof: Let $${f_p(n)={1\over {n^p}}}$$ for $n\in\mbox{{\bf Z}}_{\geq 1}$. Then for all $n\in\mbox{{\bf Z}}_{\geq 1}$ and all $p \geq 0$,

$$\begin{eqnarray*} (\sum f_p)(n)&\leq&(\sum f_p)(2n+1)\\ &=&{1\over {1^p}}+{1\ov... ...p}}+\cdots + {1\over {n^p}}\right)\\ &=&1+2^{1-p}(\sum f_p)(n). \end{eqnarray*}$$

Hence,

$$(1-2^{1-p})\left((\sum f_p)(n)\right)\leq 1.$$

If $p>1$, then $1-p<0$, so $2^{1-p}<1$ and $1-2^{1-p}$ is positive. Hence $${(\sum f_p)(n)\leq {1\over {1-2^{1-p}}}}$$; i.e., the sequence $\sum f_p$ is bounded. It is also increasing, so it converges.

If $p<1$, then $${ {1\over {n^p}}\geq{1\over n}}$$, so by using the comparison test with the harmonic series, $f_p$ is not summable. $\mid\!\mid\!\mid$

11.14   Remark. For $p>1$, the proof of the previous theorem shows that

$$\sum_{j=1}^\infty {1\over {n^p}}=\lim\sum f_p\leq{1\over {1-2^{1-p}}}.$$

Hence, we get

$$\sum_{n=1}^\infty {1\over {n^2}}\leq {1\over {1-2^{-1}}}=2,$$

and

$$\sum_{n=1}^\infty{1\over {n^4}}\leq {1\over {1-2^{-3}}}={8\over 7}=1.1428\cdots.$$

The exact values of the series (found by Euler) are

$$\sum_{n=1}^\infty{1\over {n^2}}={{\pi^2}\over 6}=1.6449\cdots$$

and

$$\sum_{n=1}^\infty{1\over {n^4}}={{\pi^4}\over {90}}=1.0823\cdots.$$

11.15   Examples. $${\left\{{1\over {n^2+n^{1/2}}}\right\}_{\geq 1}}$$ is summable, since

$$0\leq {1\over{n^2+n^{1/2}}}\leq {1\over {n^2}}\mbox{ for all }n\in\mbox{{\bf Z}}_{\geq 1}$$

and $${\left\{{1\over {n^2}}\right\}}$$ is summable.

$${\left\{1\over {1+n^{1/2}}\right\}_{n\geq 1}}$$ is not summable since

$${1\over {1+n^{1/2}}}\geq {1\over {n^{1/2}+n^{1/2}}}={1\over 2... ...t{1\over {n^{1/2}}} \mbox{ for all }n\in\mbox{{\bf Z}}_{\geq 1}$$

and $${\left\{ {1\over {n^{1/2}}}\right\}_{n\geq 1}}$$ is not summable.

11.16   Example. Let $${w={3\over 5}+{4\over 5}i}$$, and let $z\in D(0,1)$. Then $${\left\{ {1\over {n^2\vert z-w^n\vert}}\right\}_{\geq 1}}$$ is summable.

Proof: By the reverse triangle inequality, we have for all $n\in\mbox{{\bf Z}}_{\geq 1}$

$$\vert z-w^n\vert\geq\vert w^n\vert-\vert z\vert=1-\vert z\vert>0$$

so

$$0\leq{1\over {n^2\vert z-w^n\vert}}\leq {1\over {n^2(1-\vert z\vert)}}\mbox{ for all }n\in\mbox{{\bf Z}}_{\geq 1}.$$

Since $${\left\{c\cdot {1\over {n^2}}\right\}_{n\geq 1}}$$ is a summable sequence for all $c\in\mbox{{\bf C}}$, it follows from the comparison test that $${\left\{ {1\over {n^2\vert z-w^n\vert}}\right\}_{n\geq 1}}$$ is summable. $\mid\!\mid\!\mid$

11.17   Example. Let $${f(n)={{(99.99)^n}\over {n!}}}$$ for all $n\in\mbox{{\bf N}}$. Then

$$f(n+1)={{(99.99)^{n+1}}\over {(n+1)!}}={{(99.99)\cdot(99.99)^n}\over {(n+1)n!}}={{99.99}\over {n+1}}\cdot f(n).$$

If $n\geq 100$, then $n+1\geq 101$, so

$$f(n+1)={{99.99}\over {n+1}}\cdot f(n)\leq {{99.99}\over {101}}f(n).$$

Hence,

$$\begin{eqnarray*} f(101)&\leq&\left({{99.99}\over {101}}\right)f(100)\\ f(102)&... ...ht)\cdot f(102)\leq \left({{99.99}\over {100}}\right)^3 f(100). \end{eqnarray*}$$

Hence, (by induction)

$$\begin{eqnarray*} f(100+n)&\leq&\left({{99.99}\over {101}}\right)^n f(100)\\ &=... ...}f(100)\right] \\ &=&C\left({{99.99}\over {101}}\right)^{100+n} \end{eqnarray*}$$

where $${C=\left({{101}\over {99.99}}\right)^{100}f(100)}$$; i.e., $${f(j)\leq C \left({{99.99}\over {101}}\right)^j}$$ for all $j>100$. Since the geometric series $${\left\{\sum_{j=0}^n\left({{99.99}\over {101}}\right)^j\right\}}$$ converges, it follows from the comparison test that $${\left\{\sum_{j=0}^n{{(99.99)^j}\over {j!}}\right\}}$$ converges also.

11.18   Exercise. Determine whether or not the sequences below are summable:

(a) $\{(-1)^n\}$

(b) $\{(-1)^n+(-1)^{n+1}\}$

(c) $\{(-1)^n\}+\{(-1)^{n+1}\}$

(d) $${ \left\{ {{n^2}\over {n^4+1}}\right\}_{n\geq 1}}$$

(e) $${\left\{1-{n\over {n+1}}\right\}_{n\geq 1}}$$

(f) $${\left\{{1\over {n^3+\sqrt n}}\right\}_{n\geq 1}}$$

(g) $${\left\{{{n^2+n}\over {n^4+1}}\right\}_{n\geq 1}}$$

(h) $${\left\{ {{3^n}\over {n!}}\right\}}$$

11.19   Exercise. A Give examples of the following, or explain why no such examples exist.

a)

Two real sequences $f$ and $g$ such that $f$ and $g$ are not summable, but $f+g$ is summable.

b)

Two real sequences $f$ and $g$ such that $f$ and $g$ are summable, but $f+g$ is not summable.

c)

Two real sequences $f$ and $g$ such that $f(n)<g(n)$ for all $n\in\mbox{{\bf N}}$, and $g$ is summable but $f$ is not summable.

11.20   Theorem (Limit comparison test.) Let $f,g$ be sequences of positive numbers. Suppose that $${{f\over g}}$$ converges to a non-zero limit $L$. Then $f$ is summable if and only if $g$ is summable.

Proof: We know that $L>0$. Let $${N=N_{{f\over g}-\tilde L}}$$ be a precision function for $${{f\over g}-\tilde L}$$. Then

$$\left\vert {{f(n)}\over {g(n)}}-L\right\vert\leq {L\over 2}\mbox{ for all } n\geq N\left({L\over 2}\right);$$

i.e.,

$${L\over 2}\leq {{f(n)}\over {g(n)}}\leq {{3L}\over 2}\mbox{ for all } n\geq N\left({L\over 2}\right).$$

If $g$ is summable, then $${{{3L}\over 2}g}$$ is summable, and since $${f(n)\leq {{3L}\over 2}g(n)}$$ for all $${n\geq N\left({L\over 2}\right)}$$, it follows from the comparison test that $f$ is summable. If $g$ is not summable, then since $g(n) \leq {2\over L}f(n)$ for all $n \geq N({l\over 2})$ it follows that $${{2\over L}f}$$ is not summable, and hence $f$ is not summable. $\mid\!\mid\!\mid$

11.21   Example. Is $${ \left\{ {{n^2+5n+1}\over {6n^3+3n-2}}\right\}_{n\geq 1}}$$ summable? Let $${a_n={{n^2+5n+1}\over {6n^3+3n-2}}}$$. Note that $a_n>0$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. For large $n$, $a_n$ is `` like" $${{{n^2}\over {6n^3}}={1\over {6n}}}$$, so I'll compare this series with $${ \left\{ {1\over n}\right\}_{n\geq 1}}$$. Let $${b_n={1\over n}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Then

$${{a_n}\over {b_n}}={{n^3+5n^2+n}\over {6n^3+3n-2}}={{1+{5\over n}+{1\over {n^3}}}\over {6+{3\over n^2}-{2\over {n^3}}}},$$

so

$$\left\{ {{a_n}\over {b_n}}\right\}_{n\geq 1}=\left\{ {{1+{5\o... ...}\right\}_{n\geq 1}\to {{1+0+0}\over {6+0+0}}={1\over 6}\neq 0.$$

Since $${\{b_n\}_{n\geq 1}=\left\{ {1\over n}\right\}_{n\geq 1}}$$ is not summable, $${ \left\{ {{n^2+5n+1}\over {6n^3+3n-2}}\right\}_{n\geq 1}}$$ is also not summable.

11.22   Exercise. Determine whether or not the sequences below are summable.

a)

$${\left\{ {n^2+3n + 2 \over n^4 + n + 1}\right\}_{n\geq 1}}$$.

b)

$${\left\{ {n + n^2 \over n^3+n+1}\right\}_{n\geq 1}}$$.

11.23   Theorem (Ratio test.) Let $\{a_n\}$ be a sequence of positive numbers. Suppose the $${\left\{ {{a_{n+1}}\over {a_n}}\right\}}$$ converges, and $${\lim\left\{ {{a_{n+1}}\over {a_n}}\right\}=R}$$. Then, if $R<1$, $\{a_n\}$ is summable. If $R>1$, $\{a_n\}$ is not summable. (If $R=1$, the theorem makes no assertion.)

Proof: Suppose $${ \left\{ {{a_{n+1}}\over {a_n}}\right\}\to R}$$.

Case 1: $R<1$. Let $N$ be a precision function for $${\left\{{{a_{n+1}}\over {a_n}}-R\right\}}$$. Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N\left({{1-R}\over 2}\right)&\mbox{$\Longrightarrow$}&\l... ...arrow$}&{{a_{n+1}}\over {a_n}}<R+{{1-R}\over 2}= {{R+1}\over 2}. \end{eqnarray*}$$

Write $${M=N\left( {{1-R}\over 2}\right)}$$ and $${S={{1+R}\over 2}}$$, so $(0<S<1)$. Then

$$n\geq M\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}a_{n+1}\leq S\cdot a_n,$$

so

$$\begin{eqnarray*} a_M&\leq& S^0 a_M \\ a_{M+1}&\leq&Sa_M \\ a_{M+2}&\leq&S\cdot a_{M+1}\leq S^2a_M\\ a_{M+3}&\leq&S\cdot a_{M+2}\leq S^3a_M, \end{eqnarray*}$$

and (by an induction argument which I omit)

$$a_{M+k}\leq S^ka_M\mbox{ for all } k\in\mbox{{\bf N}},$$

or

$$a_{M+k}\leq S^{M+k}(a_MS^{-M})\mbox{ for all } k\in\mbox{{\bf N}},$$

or

$$a_n\leq S^n(a_MS^{-M})\mbox{ for all } n\in\mbox{{\bf Z}}_{\geq M}.$$

Since $\{S^n\}$ is a summable geometric series, it follows from the comparison test that $\{a_n\}$ is also summable.

Case 2: $(R>1)$. As before, let $N$ be a precision function for $${\left\{{{a_{n+1}}\over {a_n}}-R\right\}}$$. Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N\left({{R-1}\over 2}\right)&\mbox{$\Longrightarrow$}&\l... ...ht)={{R+1}\over 2}>1\\ &\mbox{$\Longrightarrow$}&a_{n+1} > a_n. \end{eqnarray*}$$

Hence $\{a_n\}$ is not a null sequence. So $\{a_n\}$ is not summable. $\mid\!\mid\!\mid$

11.24   Warning. The ratio test does not say that if $${ {{a_{n+1}}\over {a_n}}<1}$$ for all $n$, then $\{a_n\}$ is summable. If $${a_n={1\over n}}$$ for $n\in\mbox{{\bf Z}}_{\geq 1}$, then $${ {{a_{n+1}}\over {a_n}}={n\over {n+1}}<1}$$ for all $n$ but $\{a_n\}$ is not summable. (In this case, $${\lim\left\{ {{a_{n+1}}\over {a_n}}\right\}=1}$$, and the ratio test does not apply.)

If $${b_n={1\over {n^2}}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$, then $${ {{b_{n+1}}\over {b_n}}=\left( {{n^2}\over {(n+1)^2}}\right)}$$ for all $n$ and hence
$${\lim\left\{ {{b_{n+1}}\over {b_n}}\right\}=1}$$, and $\{b_n\}$ is summable. These examples show that when $${\lim\left\{ {{a_{n+1}}\over {a_n}}\right\}=1}$$ the ratio test gives no useful information.

11.25   Remark. If, in applying the ratio test, you find that $${ {{a_{n+1}}\over {a_n}}\geq 1}$$ for all large $n$, you can conclude that $\sum\{a_n\}$ diverges (even if $${\lim\left\{ {{a_{n+1}}\over {a_n}}\right\}}$$ does not exist), since this condition shows that $\{a_n\}$ is not a null sequence.

11.26   Example. Let $t$ be a positive number and let $${a_n={{(3n)!t^n}\over {(n!)^3}}}$$. We apply the ratio test to the series $\sum\{a_n\}$.

$${{a_{n+1}}\over {a_n}}={{\left(3(n+1)\right)!t^{n+1}(n!)^3}\over {[(n+1)!]^3(3n)!t^n}}.$$

Note that

$$\begin{eqnarray*} \left(3(n+1)\right)!&=&(3n+3)!=(3n+3)(3n+2)!=(3n+3)(3n+2)(3n+1)! \\ &=&(3n+3)(3n+2)(3n+1)(3n)!. \end{eqnarray*}$$

Hence

$$\begin{eqnarray*} {{a_{n+1}}\over {a_n}}&=&{{(3n)!t\cdot(3n+3)(3n+2)(3n+1)}\over... ...r n}}}\right)\left( {{3+{1\over n}}\over {1+{1\over n}}}\right). \end{eqnarray*}$$

From this we see that $${\Big\{ {a_{n+1}\over a_n}\Big\} \to 27t}$$. The ratio test says that if $27t<1$ (i.e., if ${t<{1\over {27}}}$), then $${\left\{ {{(3n)!t^n}\over {(n!)^3}}\right\}}$$ is summable, and if ${t>{1\over {27}}}$, then the sequence is not summable.

Can we figure out what happens in the case $${t={1\over {27}}}$$? For $${t={1\over {27}}}$$, our formula above gives us

$${{a_{n+1}}\over {a_n}}={{\left(1+{2\over {3n}}\right)\left({1... ... n}\right)}\over {\left(1+{1\over n}\right)^2}}={n\over {n+1}};$$

i.e., $${a_{n+1}>{n\over {n+1}}a_n}$$ for $n\geq 1$. Thus,

$$\begin{eqnarray*} a_2&\geq&{1\over 2}\cdot a_1\\ a_3&\geq&{2\over 3}a_2\geq{2\o... ...\geq&{4\over 5}a_4\geq{4\over 5}\cdot{1\over 4}a_1={1\over 5}a_1 \end{eqnarray*}$$

and (by induction),

$$a_n\geq{1\over n}a_1\mbox{ for }n\geq 1.$$

Since $${ \left\{{{a_1}\over n}\right\}_{n\geq 1}}$$ is not summable, it follows that $\{a_n\}$ is not summable for $${t={1\over {27}}}$$.

11.27   Example. Let $${b_n={{(n!)^24^n}\over {(2n)!}}}$$ for all $n\in\mbox{{\bf N}}$. I'll apply ratio test to $\sum\{b_n\}$. For all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} {{b_{n+1}}\over {b_n}}&=&{{(n+1)!^24^{n+1}}\over {(2n+2)!}}\cd... ...)}}={{2n+2}\over {2n+1}}={{1+{1\over n}}\over {1+{1\over {2n}}}} \end{eqnarray*}$$

Hence $${\left\{ {{b_{n+1}}\over {b_n}}\right\}\to 1}$$ and the ratio test does not apply. But since $${ {{2n+2}\over {2n+1}}>1}$$ for all $n$, I conclude that $\{b_n\}$ is an increasing sequence and hence $\sum\{b_n\}$ diverges.

11.28   Exercise. For each of the series below, A determine for which $x\in[0,\infty)$ the series converges.

a) $${\sum\left\{ {{x^n}\over {n!}}\right\}}$$

b) $${\sum\left\{ {{x^{2n}}\over {(2n)!}}\right\}}$$

c) $${\sum\left\{ {{3^nx^n}\over {n^2}}\right\}_{n\geq 1}}$$

d) $${\sum\left\{ {{x^n}\over {2^n\sqrt n}}\right\}_{n\geq 1}}$$

e) $\sum\{nx^n\}_{n\geq 1}$

f) $\sum\{n!x^n\}$

g) $${\sum\left\{ {{(n!)^2x^n}\over {(2n)!}}\right\}}$$ [For this series, there is one $x\in[0,\infty)$ for which you don't need to answer the question.]