Comparison Tests

${\displaystyle \sum }_{n=1}^{\infty }{a}_n$

where $a_n=\frac{2}{n\left(n+1\right)}$

${\displaystyle \sum }_{n=1}^{\infty }{a}_n$

where $a_n=\frac{1}{n\left(n+1\text{/}2\right)}$

Converges by comparison with $1\text{/}{n}^2.$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2\left(n+1\right)}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2n-1}$

Diverges by comparison with harmonic series, since $2n-1\ge n.$

${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{{\left(n\text{ln}n\right)}^2}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{n\text{!}}{\left(n+2\right)\text{!}}$

$a_n=1\text{/}\left(n+1\right)\left(n+2\right)<1\text{/}{n}^2.$

Converges by comparison with p-series, $p=2.$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n\text{!}}$

${\displaystyle \sum _{n=1}^{\infty }\frac{\text{sin}\left(1\text{/}n\right)}{n}}$

$\text{sin}\left(1\text{/}n\right)\le 1\text{/}n,$

so converges by comparison with p-series, $p=2.$

${\displaystyle \sum _{n=1}^{\infty }\frac{{\text{sin}}^{2}n}{n^2}}$

${\displaystyle \sum _{n=1}^{\infty }\frac{\text{sin}\left(1\text{/}n\right)}{\sqrt{n}}}$

$\text{sin}\left(1\text{/}n\right)\le 1,$

so converges by comparison with p-series, **$p=3\text{/}2.$

${\displaystyle \sum }_{n=1}^{\infty }\frac{n^{1.2}-1}{n^{2.3}+1}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{\sqrt{n+1}-\sqrt{n}}{n}$

Since $\sqrt{n+1}-\sqrt{n}=1\text{/}\left(\sqrt{n+1}+\sqrt{n}\right)\le 2\text{/}\sqrt{n},$

series converges by comparison with p-series **for $p=\mathrm{1.5.}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{\sqrt[4]{n}}{\sqrt[3]{n^4+n^2}}$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\frac{\text{ln}n}{n}\right)}^2$

Converges by limit comparison with p-series **for $p>1.$

${\displaystyle \sum }_{n=1}^{\infty }{\left(\frac{\text{ln}n}{n^{0.6}}\right)}^2$

${\displaystyle \sum }_{n=1}^{\infty }\frac{\text{ln}\left(1+\frac{1}{n}\right)}{n}$

Converges by limit comparison with p-series, **$p=2.$

${\displaystyle \sum }_{n=1}^{\infty }\text{ln}\left(1+\frac{1}{n^2}\right)$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{4^n-3^n}$

Converges by limit comparison with $4^{\text{−}n}.$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n^2-n\text{sin}n}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{e^{\left(1.1\right)n}-3^n}$

Converges by limit comparison with $1\text{/}{e}^{1.1n}.$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{e^{\left(1.01\right)n}-3^n}$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n^{1+1\text{/}n}}$

Diverges by limit comparison with harmonic series.

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2^{1+1\text{/}n}{n}^{1+1\text{/}n}}$

${\displaystyle \sum }_{n=1}^{\infty }\left(\frac{1}{n}-\text{sin}\left(\frac{1}{n}\right)\right)$

Converges by limit comparison with p-series, $p=3.$

${\displaystyle \sum }_{n=1}^{\infty }\left(1-\text{cos}\left(\frac{1}{n}\right)\right)$

${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n}\left({\text{tan}}^{\mathrm{-1}}n-\frac{\pi }{2}\right)$

Converges by limit comparison with p-series, $p=3.$

${\displaystyle \sum }_{n=1}^{\infty }{\left(1-\frac{1}{n}\right)}^{n.n}$

(Hint:${\left(1-\frac{1}{n}\right)}^n\to 1\text{/}e.)$

${\displaystyle \sum }_{n=1}^{\infty }\left(1-e^{\mathrm{-1}\text{/}n}\right)$

(Hint:$1\text{/}e\approx {\left(1-1\text{/}n\right)}^n,$

so $1-e^{\mathrm{-1}\text{/}n}\approx 1\text{/}n.)$

Diverges by limit comparison with $1\text{/}n.$

Does ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{{\left(\text{ln}n\right)}^p}$

converge if $p$

is large enough? If so, for which $p\text{?}$

Does ${\displaystyle \sum }_{n=1}^{\infty }{\left(\frac{\left(\text{ln}n\right)}{n}\right)}^p$

converge if $p$

is large enough? If so, for which $p\text{?}$

Converges for $p>1$

by comparison with a $p$

series for slightly smaller $p.$

For which $p$

does the series ${\displaystyle \sum }_{n=1}^{\infty }{2}^{pn}\text{/}{3}^n$

converge?

For which $p>0$

does the series ${\displaystyle \sum }_{n=1}^{\infty }\frac{n^p}{2^n}$

converge?

Converges for all $p>0.$

For which $r>0$

does the series ${\displaystyle \sum }_{n=1}^{\infty }\frac{r^{n^2}}{2^n}$

converge?

For which $r>0$

does the series ${\displaystyle \sum }_{n=1}^{\infty }\frac{2^n}{r^{n^2}}$

converge?

Converges for all $r>1.$

If $r>1$

then $r^n>4,$

say, once $n>\text{ln}\left(2\right)\text{/}\text{ln}\left(r\right)$

and then the series converges by limit comparison with a geometric series with ratio $1\text{/}2.$

Find all values of $p$

and $q$

such that ${\displaystyle \sum }_{n=1}^{\infty }\frac{n^p}{{\left(n\text{!}\right)}^q}$

converges.

Does ${\displaystyle \sum }_{n=1}^{\infty }\frac{{\text{sin}}^2\left(nr\text{/}2\right)}{n}$

converge or diverge? Explain.

The numerator is equal to $1$

when $n$

is odd and $0$

when $n$

is even, so the series can be rewritten ${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2n+1},$

which diverges by limit comparison with the harmonic series.

Explain why, for each $n,$

at least one of $\{\backslash |\text{sin}n\backslash |,\backslash |\text{sin}\left(n+1\right)\backslash |\text{,...},\backslash |\text{sin}n+6\backslash |\}$

is larger than $1\text{/}2.$

Use this relation to test convergence of ${\displaystyle \sum }_{n=1}^{\infty }\frac{\backslash |\text{sin}n\backslash |}{\sqrt{n}}.$

Suppose that $a_n\ge 0$

and $b_n\ge 0$

and that ${\displaystyle \sum _{n=1}^{\infty }{a}^2{}_n}$

and ${\displaystyle \sum _{n=1}^{\infty }{b}^2{}_n}$

converge. Prove that ${\displaystyle \sum _{n=1}^{\infty }{a}_n{b}_n}$

converges and ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{b}_n\le \frac{1}{2}\left({\displaystyle \sum _{n=1}^{\infty }{a}_n^2}+{\displaystyle \sum _{n=1}^{\infty }{b}_n^2}\right).$

${\left(a-b\right)}^2=a^2-2ab+b^2$

or $a^2+b^2\ge 2ab,$

so convergence follows from comparison of $2a_n{b}_n$

with $a^2{}_n+b^2{}_n.$

Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.

Does ${\displaystyle \sum _{n=1}^{\infty }{2}^{\text{−}\text{ln}\text{ln}n}}$

converge? (Hint: Write $2^{\text{ln}\text{ln}n}$

as a power of $\text{ln}n.)$

Does ${\displaystyle \sum _{n=1}^{\infty }{\left(\text{ln}n\right)}^{\text{−}\text{ln}n}}$

converge? (Hint: Use $t=e^{\text{ln}\left(t\right)}$

to compare to a $p-\text{series}\text{.})$

${\left(\text{ln}n\right)}^{\text{−}\text{ln}n}=e^{\text{−}\text{ln}\left(n\right)\text{ln}\text{ln}\left(n\right)}.$

If $n$

is sufficiently large, then $\text{ln}\text{ln}n>2,$

so ${\left(\text{ln}n\right)}^{\text{−}\text{ln}n}<1\text{/}{n}^2,$

and the series converges by comparison to a $p-\text{series}\text{.}$

Does ${\displaystyle \sum _{n=2}^{\infty }{\left(\text{ln}n\right)}^{\text{−}\text{ln}\text{ln}n}}$

converge? (Hint: Compare $a_n$

to $1\text{/}n.)$

Show that if $a_n\ge 0$

and ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

converges, then ${\displaystyle \sum _{n=1}^{\infty }{a}^2{}_n}$

converges. If ${\displaystyle \sum _{n=1}^{\infty }{a}^2{}_n}$

converges, does ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

necessarily converge?

$a_n\to 0,$

so $a^2{}_n\le \backslash |a_n\backslash |$

for large $n.$

Convergence follows from limit comparison. ${\displaystyle \sum 1\text{/}{n}^2}$

converges, but ${\displaystyle \sum 1\text{/}n}$

does not, so the fact that ${\displaystyle \sum _{n=1}^{\infty }{a}^2{}_n}$

converges does not imply that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

converges.

Suppose that $a_n>0$

for all $n$

and that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

converges. Suppose that $b_n$

is an arbitrary sequence of zeros and ones. Does ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{b}_n$

necessarily converge?

Suppose that $a_n>0$

for all $n$

and that ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

diverges. Suppose that $b_n$

is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does ${\displaystyle \sum _{n=1}^{\infty }{a}_n}{b}_n$

necessarily diverge?

No. ${\displaystyle \sum _{n=1}^{\infty }1\text{/}n}$

diverges. Let $b_k=0$

unless $k=n^2$

for some $n.$

Then ${\displaystyle \sum _k{b}_k\text{/}k}={\displaystyle \sum 1\text{/}{k}^2}$

converges.

Complete the details of the following argument: If ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$

converges to a finite sum $s,$

then $\frac{1}{2}s=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\text{⋯}$

and $s-\frac{1}{2}s=1+\frac{1}{3}+\frac{1}{5}+\text{⋯}.$

Why does this lead to a contradiction?

Show that if $a_n\ge 0$

and ${\displaystyle \sum _{n=1}^{\infty }{a}^2{}_n}$

converges, then ${\displaystyle \sum _{n=1}^{\infty }{\text{sin}}^2\left(a_n\right)}$

converges.

$\backslash |\text{sin}t\backslash |\le \backslash |t\backslash |,$

so the result follows from the comparison test.

Suppose that $a_n\text{/}{b}_n\to 0$

in the comparison test, where $a_n\ge 0$

and $b_n\ge 0.$

Prove that if ${\displaystyle \sum b_n}$

converges, then ${\displaystyle \sum a_n}$

converges.

Let $b_n$

be an infinite sequence of zeros and ones. What is the largest possible value of $x={\displaystyle \sum _{n=1}^{\infty }{b}_n\text{/}{2}^n}\text{?}$

By the comparison test, $x={\displaystyle \sum _{n=1}^{\infty }{b}_n\text{/}{2}^n}\le {\displaystyle \sum _{n=1}^{\infty }1\text{/}{2}^n}=1.$

Let $d_n$

be an infinite sequence of digits, meaning $d_n$

takes values in $\{0,1\text{,…},9\}.$

What is the largest possible value of $x={\displaystyle \sum _{n=1}^{\infty }{d}_n\text{/}{10}^n}$

that converges?

Explain why, if $x>1\text{/}2,$

then $x$

cannot be written $x={\displaystyle \sum _{n=2}^{\infty }\frac{b_n}{2^n}}\left(b_n=0\text{or}1,b_1=0\right).$

If $b_1=0,$

then, by comparison, $x\le {\displaystyle \sum _{n=2}^{\infty }1\text{/}{2}^n}=1\text{/}2.$

[T] Evelyn has a perfect balancing scale, an unlimited number of $1\text{-kg}$

weights, and one each of $1\text{/}2\text{-kg},1\text{/}4\text{-kg},1\text{/}8\text{-kg},$

and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1\text{-kg}$

weights, and nine each of $0.1\text{-kg,}$

$0.01\text{-kg},0.001\text{-kg,}$

and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

Yes. Keep adding $1\text{-kg}$

weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the $1\text{-kg}$

weights, and add $0.1\text{-kg}$

weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last $0.1\text{-kg}$

weight. Start adding $0.01\text{-kg}$

weights. If it balances, stop. If it tips to the side with the weights, remove the last $0.01\text{-kg}$

weight that was added. Continue in this way for the $0.001\text{-kg}$

weights, and so on. After a finite number of steps, one has a finite series of the form $A+{\displaystyle \sum _{n=1}^N{s}_n}\text{/}{10}^n$

where $A$

is the number of full kg weights and $d_n$

is the number of $1\text{/}{10}^n\text{-kg}$

weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the $N\text{th}$

partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most $1\text{/}{10}^N.$

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{2n}}$

is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which $n$

is odd. Let $m>1$

be fixed. Show, more generally, that deleting all terms $1\text{/}n$

where $n=mk$

for some integer $k$

also results in a divergent series.

In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$

by removing any term $1\text{/}n$

if a given digit, say $9,$

appears in the decimal expansion of $n.$

Argue that this depleted harmonic series converges by answering the following questions.

1. How many whole numbers $n$

   have

   $d$

   digits?

2. How many $d\text{-digit}$

   whole numbers

   $h\left(d\right).$

   do not contain

   $9$

   as one or more of their digits?

3. What is the smallest $d\text{-digit}$

   number

   $m\left(d\right)\text{?}$
4. Explain why the deleted harmonic series is bounded by ${\displaystyle \sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}}.$
5. Show that ${\displaystyle \sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}}$

   converges.

a. ${10}^d-{10}^{d-1}<{10}^d$

b. $h\left(d\right)<9^d$

c. $m\left(d\right)={10}^{d-1}+1$

d. Group the terms in the deleted harmonic series together by number of digits. $h\left(d\right)$

bounds the number of terms, and each term is at most $1\text{/}m\left(d\right).$

${\displaystyle \sum _{d=1}^{\infty }h\left(d\right)\text{/}m\left(d\right)}\le {\displaystyle \sum _{d=1}^{\infty }{9}^d\text{/}{\left(10\right)}^{d-1}}\le 90.$

One can actually use comparison to estimate the value to smaller than $80.$

The actual value is smaller than $23.$

Suppose that a sequence of numbers $a_n>0$

has the property that $a_1=1$

and $a_{n+1}=\frac{1}{n+1}{S}_n,$

where $S_n=a_1+\text{⋯}+a_n.$

Can you determine whether ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

converges? (Hint: $S_n$

is monotone.)

Suppose that a sequence of numbers $a_n>0$

has the property that $a_1=1$

and $a_{n+1}=\frac{1}{{\left(n+1\right)}^2}{S}_n,$

where $S_n=a_1+\text{⋯}+a_n.$

Can you determine whether ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$

converges? (Hint: $S_2=a_2+a_1=a_2+S_1=a_2+1=1+1\text{/}4=\left(1+1\text{/}4\right)S_1,$

$S_3=\frac{1}{3^2}{S}_2+S_2=\left(1+1\text{/}9\right)S_2=\left(1+1\text{/}9\right)\left(1+1\text{/}4\right)S_1,$

etc. Look at $\text{ln}\left(S_n\right),$

and use $\text{ln}\left(1+t\right)\le t,$

$t>0.)$

Continuing the hint gives $S_N=\left(1+1\text{/}{N}^2\right)\left(1+1\text{/}{\left(N-1\right)}^2\text{…}\left(1+1\text{/}4\right)\right).$

Then $\text{ln}\left(S_N\right)=\text{ln}\left(1+1\text{/}{N}^2\right)+\text{ln}\left(1+1\text{/}{\left(N-1\right)}^2\right)+\text{⋯}+\text{ln}\left(1+1\text{/}4\right).$

Since $\text{ln}\left(1+t\right)$

is bounded by a constant times $t,$

when $0<t<1$

one has $\text{ln}\left(S_N\right)\le C{\displaystyle \sum _{n=1}^N\frac{1}{n^2}},$

which converges by comparison to the p-series for $p=2.$