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Infinitely Small Sequence

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Definition. Sequence xn{x}_{{n}} is called infinitesimal if its limit is 0.

In other words, if for every ϵ>0\epsilon>{0} there is number N=Nϵ{N}={N}_{{\epsilon}}, such that for n>N{n}>{N}, ∣xn−0∣=∣xn∣<ϵ.{\left|{x}_{{n}}-{0}\right|}={\left|{x}_{{n}}\right|}<\epsilon.

We can reformulate definition as follows: sequence xn{x}_{{n}} is infinitesimal if its absolute value becomes less than some specified ϵ>0\epsilon>{0}, starting with some number.

If we now return to the variant xn{x}_{{n}} that has limit a{a}, then difference αn=xn−a\alpha_{{n}}={x}_{{n}}-{a} will be infinitesimal, because by definition of limit ∣αn∣=∣xn−a∣<ϵ{\left|\alpha_{{n}}\right|}={\left|{x}_{{n}}-{a}\right|}<\epsilon. And vice versa, if αn\alpha_{{n}} is infinitesimal then xn→a{x}_{{n}}\to{a}.

This leads to the following fact.

Fact. xn{x}_{{n}} has limit a{a} if and only if αn=xn−a\alpha_{{n}}={x}_{{n}}-{a} is infinitesimal.

Therefore, if xn→a{x}_{{n}}\to{a} then xn{x}_{{n}} can be represented as xn=a+αn{x}_{{n}}={a}+\alpha_{{n}}.

This fact is often useful for finding limits.

Example 1. Consider sequences xn=1n{x}_{{n}}=\frac{{1}}{{n}}, xn=−1n{x}_{{n}}=-\frac{{1}}{{n}}, xn=(−1)n+1n{x}_{{n}}=\frac{{{{\left(-{1}\right)}}^{{{n}+{1}}}}}{{n}}.

Corresponding lists are

{1,12,13,14,…}{\left\{{1},\frac{{1}}{{2}},\frac{{1}}{{3}},\frac{{1}}{{4}},\ldots\right\}},

{−1,−12,−13,−14,…}{\left\{-{1},-\frac{{1}}{{2}},-\frac{{1}}{{3}},-\frac{{1}}{{4}},\ldots\right\}},

{1,−12,13,−14,…}{\left\{{1},-\frac{{1}}{{2}},\frac{{1}}{{3}},-\frac{{1}}{{4}},\ldots\right\}}.

All variants are infinitesimal because ∣xn−0∣=∣xn∣=∣1n∣<ϵ{\left|{x}_{{n}}-{0}\right|}={\left|{x}_{{n}}\right|}={\left|\frac{{1}}{{n}}\right|}<\epsilon when n>1ϵ{n}>\frac{{1}}{\epsilon}. Therefore, we can take (recall that N{N} is natural number) N=Nϵ>[1ϵ]{N}={N}_{\epsilon}>{\left[\frac{{1}}{\epsilon}\right]}, where [x]{\left[{x}\right]} is a floor function.

You see that they are infinitesimal (so their limit is 0), but they behave differently: first is always greater 0, second is always less than 0, third alternates sign.

Example 2. Consider sequence xn=2+(−1)nn{x}_{{n}}=\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{n}}.

Corresponding list is {1,32,13,34,15,12,…}{\left\{{1},\frac{{3}}{{2}},\frac{{1}}{{3}},\frac{{3}}{{4}},\frac{{1}}{{5}},\frac{{1}}{{2}},\ldots\right\}}.

By triangle inequality we have that ∣xn∣=∣2n+(−1)nn∣≤∣2n∣+∣(−1)nn∣=∣2n∣+∣1n∣=∣3n∣{\left|{x}_{{n}}\right|}={\left|\frac{{2}}{{n}}+\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}\right|}\le{\left|\frac{{2}}{{n}}\right|}+{\left|\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}\right|}={\left|\frac{{2}}{{n}}\right|}+{\left|\frac{{1}}{{n}}\right|}={\left|\frac{{3}}{{n}}\right|}.

Therefore, ∣3n∣<ϵ{\left|\frac{{3}}{{n}}\right|}<\epsilon when n>3ϵ{n}>\frac{{3}}{\epsilon}. So, Nϵ=[3ϵ]{N}_{\epsilon}={\left[\frac{{3}}{\epsilon}\right]}. Thus, this variant is infinitesimal.

Note, that here we see different behavior comparing with example 1: variant alternately approach limit 0 and move away from it.

Example 3. Consider sequence xn=1+(−1)nn{x}_{{n}}=\frac{{{1}+{{\left(-{1}\right)}}^{{n}}}}{{n}}.

Corresponding sequence is {0,1,0,12,…}{\left\{{0},{1},{0},\frac{{1}}{{2}},\ldots\right\}}.

By triangle inequality we have that ∣xn∣=∣1n+(−1)nn∣≤∣1n∣+∣(−1)nn∣=∣1n∣+∣1n∣=∣2n∣{\left|{x}_{{n}}\right|}={\left|\frac{{1}}{{n}}+\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}\right|}\le{\left|\frac{{1}}{{n}}\right|}+{\left|\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}\right|}={\left|\frac{{1}}{{n}}\right|}+{\left|\frac{{1}}{{n}}\right|}={\left|\frac{{2}}{{n}}\right|}.

Therefore, ∣2n∣<ϵ{\left|\frac{{2}}{{n}}\right|}<\epsilon when n>2ϵ{n}>\frac{{2}}{\epsilon}. So, Nϵ=[2ϵ]{N}_{\epsilon}={\left[\frac{{2}}{\epsilon}\right]}. Thus, this variant is infinitesimal, in other word its limit is 0.

Note that sequence takes limiting value.

These simple examples shows that although all four sequences have same limit, but they approach it differently. First three sequences don't take limiting value, while fourth does.

The only thing that matters is that difference between values in sequence and limit should be infinitesimal for all values that lie sufficiently far.

Example 4. Let's take slighly harder example. Consider sequence xn=n2−n+23n2+2n−4{x}_{{n}}=\frac{{{{n}}^{{2}}-{n}+{2}}}{{{3}{{n}}^{{2}}+{2}{n}-{4}}}. Show that its limit is 13\frac{{1}}{{3}}.

To show that limit of this sequence is 13\frac{{1}}{{3}} we need to show that ∣xn−13∣{\left|{x}_{{n}}-\frac{{1}}{{3}}\right|} is infinitesimal.

So, after algebraic manipulations we obtain that ∣xn−13∣=∣n2−n+23n2+2n−4−13∣=∣−5n−103(3n2+2n−4)∣=5n−103(3n2+2n−4)<5n3(3n2−4){\left|{x}_{{n}}-\frac{{1}}{{3}}\right|}={\left|\frac{{{{n}}^{{2}}-{n}+{2}}}{{{3}{{n}}^{{2}}+{2}{n}-{4}}}-\frac{{1}}{{3}}\right|}={\left|-\frac{{{5}{n}-{10}}}{{{3}{\left({3}{{n}}^{{2}}+{2}{n}-{4}\right)}}}\right|}=\frac{{{5}{n}-{10}}}{{{3}{\left({3}{{n}}^{{2}}+{2}{n}-{4}\right)}}}<\frac{{{5}{n}}}{{{3}{\left({3}{{n}}^{{2}}-{4}\right)}}}.

Since n{n} is large then n2−4>0{{n}}^{{2}}-{4}>{0}, so 3n2−4=2n2+n2−4>2n2{3}{{n}}^{{2}}-{4}={2}{{n}}^{{2}}+{{n}}^{{2}}-{4}>{2}{{n}}^{{2}}.

So, 5n3(3n2−4)<5n3⋅2n2=56n<1n\frac{{{5}{n}}}{{{3}{\left({3}{{n}}^{{2}}-{4}\right)}}}<\frac{{{5}{n}}}{{{3}\cdot{2}{{n}}^{{2}}}}=\frac{{5}}{{{6}{n}}}<\frac{{1}}{{n}}.

Therefore, ∣xn−13∣<ϵ{\left|{x}_{{n}}-\frac{{1}}{{3}}\right|}<\epsilon when 1n<ϵ\frac{{1}}{{n}}<\epsilon or n>1ϵ{n}>\frac{{1}}{\epsilon}. In other words if n>Nϵ=[1ϵ]{n}>{N}_{\epsilon}={\left[\frac{{1}}{\epsilon}\right]}.

Thus, xn→13{x}_{{n}}\to\frac{{1}}{{3}}.

Example 5. Let sequence is given by formula xn=a1n=an{x}_{{n}}={{a}}^{{\frac{{1}}{{n}}}}={\sqrt[{{n}}]{{{a}}}} (a>1). Let's prove that xn→1.{x}_{{n}}\to{1}.

Using Bernoulli inequality γn>1+n(γ−1){\gamma}^{{n}}>{1}+{n}{\left(\gamma-{1}\right)} where n{n} is natural and n>1{n}>{1}, γ>1\gamma>{1}, with γ=a1n\gamma={{a}}^{{\frac{{1}}{{n}}}}, we obtain that (a1n)n>1+n(a1n−1){{\left({{a}}^{{\frac{{1}}{{n}}}}\right)}}^{{n}}>{1}+{n}{\left({{a}}^{{\frac{{1}}{{n}}}}-{1}\right)} or a1n−1<a−1n{{a}}^{{\frac{{1}}{{n}}}}-{1}<\frac{{{a}-{1}}}{{n}}.

This means that ∣xn−1∣=an−1<a−1n<ϵ{\left|{x}_{{n}}-{1}\right|}={\sqrt[{{n}}]{{{a}}}}-{1}<\frac{{{a}-{1}}}{{n}}<\epsilon when n>Nϵ=[a−1ϵ]{n}>{N}_{\epsilon}={\left[\frac{{{a}-{1}}}{\epsilon}\right]}.

However, we can go another way.

Inequality ∣xn−1∣=a1n−1<ϵ{\left|{x}_{{n}}-{1}\right|}={{a}}^{{\frac{{1}}{{n}}}}-{1}<\epsilon is equivalent to the inequality 1n<log⁡a(1+ϵ)\frac{{1}}{{n}}<{\log}_{{a}}{\left({1}+\epsilon\right)} or n>1log⁡a(1+ϵ){n}>\frac{{1}}{{{\log}_{{a}}{\left({1}+\epsilon\right)}}}, so it holds when n>Nϵ=[1log⁡a(1+ϵ)]{n}>{N}_{\epsilon}={\left[\frac{{1}}{{{\log}_{{a}}{\left({1}+\epsilon\right)}}}\right]}.

As can be seen two ways of thinking led us to the two different expressions for Nϵ{N}_{\epsilon}. For example, when a=10,ϵ=0.01{a}={10},\epsilon={0.01} we obtain that N0.01=90.01=900{N}_{{{0.01}}}=\frac{{9}}{{{0.01}}}={900} according to the first way and N0.01=[10.00432….]=231{N}_{{{0.01}}}={\left[\frac{{1}}{{{0.00432}\ldots.}}\right]}={231} according to the second way. Second way gave us the smallest of all possible values for N0.01{N}_{{{0.01}}} because 101231=1.010017{{10}}^{{\frac{{1}}{{231}}}}={1.010017} is different from 1 on more than ϵ=0.01\epsilon={0.01}.

Same will be in general case, because when a≤1log⁡a(1+ϵ){a}\le\frac{{1}}{{{\log}_{{a}}{\left({1}+\epsilon\right)}}} we have that a1n−1≥ϵ{{a}}^{{\frac{{1}}{{n}}}}-{1}\ge\epsilon.

However, we are not interested in finding the smallest value of Nϵ{N}_{\epsilon} if we only want to find limit. We are interested in finding such Nϵ{N}_{\epsilon} that inequality will hold for all n≥Nϵ{n}\ge{N}_{\epsilon}. We don't care is it smallest value or not.

Example 6. Important example of infinitesimal is sequence αn=qn\alpha_{{n}}={{q}}^{{n}} where ∣q∣<1{\left|{q}\right|}<{1}.

To prove that αn→0\alpha_{{n}}\to{0} consider inequlaity ∣αn∣=∣q∣n<ϵ{\left|\alpha_{{n}}\right|}={{\left|{q}\right|}}^{{n}}<\epsilon. This inequality is equivalent to the inequality lg⁡(∣q∣n)<lg⁡(ϵ){\lg{{\left({{\left|{q}\right|}}^{{n}}\right)}}}<{\lg{{\left(\epsilon\right)}}} or nlg⁡∣q∣<lg⁡(ϵ){n}{\lg}{\left|{q}\right|}<{\lg{{\left(\epsilon\right)}}} i.e. n>lg⁡(ϵ)lg⁡∣q∣{n}>\frac{{{\lg{{\left(\epsilon\right)}}}}}{{{\lg}{\left|{q}\right|}}}.

Therefore, if we take (considering ϵ<1\epsilon<{1}) Nϵ=[lg⁡(ϵ)lg⁡∣q∣]{N}_{\epsilon}={\left[\frac{{{\lg{{\left(\epsilon\right)}}}}}{{{\lg}{\left|{q}\right|}}}\right]} then for n>Nϵ{n}>{N}_{\epsilon} above inequality will hold.

Similarly it can be shown that variant βn=A⋅qn\beta_{{n}}={A}\cdot{{q}}^{{n}}, where ∣q∣<1{\left|{q}\right|}<{1} and A is constant, is also infinitesimal.

Example 7. Now, consider infinite decreasing geometric sequence a,aq,aq2,…,aqn−1,…{a},{a}{q},{a}{{q}}^{{2}},\ldots,{a}{{q}}^{{{n}-{1}}},\ldots

where ∣q∣<1{\left|{q}\right|}<{1}. Let's find its sum.

Sum of infinite progression is limit as n→∞{n}\to\infty of partial sum of geometric progression sn{s}_{{n}}.

We have that sn=a−aqn1−q=a1−q−a1−qqn{s}_{{n}}=\frac{{{a}-{a}{{q}}^{{n}}}}{{{1}-{q}}}=\frac{{a}}{{{1}-{q}}}-\frac{{a}}{{{1}-{q}}}{{q}}^{{n}}.

This equality means that variant sn{s}_{{n}} is different from number a1−q\frac{{a}}{{{1}-{q}}} on αn=11−qqn\alpha_{{n}}=\frac{{1}}{{{1}-{q}}}{{q}}^{{n}}. But as we saw in example 6 αn\alpha_{{n}} is infinitesimal. Therefore, s=lim⁡(sn)=11−q{s}=\lim{\left({s}_{{n}}\right)}=\frac{{1}}{{{1}-{q}}}.

Therefore, when ∣q∣<1{\left|{q}\right|}<{1} we have that a+aq+aq2+…+aqn−1+…=a1−q{a}+{a}{q}+{a}{{q}}^{{2}}+\ldots+{a}{{q}}^{{{n}-{1}}}+\ldots=\frac{{a}}{{{1}-{q}}}.