# Infinitely Small Sequence

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

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

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

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

This leads to the following fact.

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

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

This fact is often useful for finding limits.

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

Corresponding lists are

${\left\{{1},\frac{{1}}{{2}},\frac{{1}}{{3}},\frac{{1}}{{4}},\ldots\right\}}$,

${\left\{-{1},-\frac{{1}}{{2}},-\frac{{1}}{{3}},-\frac{{1}}{{4}},\ldots\right\}}$,

${\left\{{1},-\frac{{1}}{{2}},\frac{{1}}{{3}},-\frac{{1}}{{4}},\ldots\right\}}$.

All variants are infinitesimal because ${\left|{x}_{{n}}-{0}\right|}={\left|{x}_{{n}}\right|}={\left|\frac{{1}}{{n}}\right|}<\epsilon$ when ${n}>\frac{{1}}{\epsilon}$. Therefore, we can take (recall that ${N}$ is natural number) ${N}={N}_{\epsilon}>{\left[\frac{{1}}{\epsilon}\right]}$, where ${\left[{x}\right]}$ 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 ${x}_{{n}}=\frac{{{2}+{{\left(-{1}\right)}}^{{n}}}}{{n}}$.

Corresponding list is ${\left\{{1},\frac{{3}}{{2}},\frac{{1}}{{3}},\frac{{3}}{{4}},\frac{{1}}{{5}},\frac{{1}}{{2}},\ldots\right\}}$.

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

Therefore, ${\left|\frac{{3}}{{n}}\right|}<\epsilon$ when ${n}>\frac{{3}}{\epsilon}$. So, ${N}_{\epsilon}={\left[\frac{{3}}{\epsilon}\right]}$. 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 ${x}_{{n}}=\frac{{{1}+{{\left(-{1}\right)}}^{{n}}}}{{n}}$.

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

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

Therefore, ${\left|\frac{{2}}{{n}}\right|}<\epsilon$ when ${n}>\frac{{2}}{\epsilon}$. So, ${N}_{\epsilon}={\left[\frac{{2}}{\epsilon}\right]}$. 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 ${x}_{{n}}=\frac{{{{n}}^{{2}}-{n}+{2}}}{{{3}{{n}}^{{2}}+{2}{n}-{4}}}$. Show that its limit is $\frac{{1}}{{3}}$.

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

So, after algebraic manipulations we obtain that ${\left|{x}_{{n}}-\frac{{1}}{{3}}\right|}={\left|\frac{{{{n}}^{{2}}-{n}+{2}}}{{{3}{{n}}^{{2}}+{2}{n}-{4}}}-\frac{{1}}{{3}}\right|}={\left|-\frac{{{5}{n}-{10}}}{{{3}{\left({3}{{n}}^{{2}}+{2}{n}-{4}\right)}}}\right|}=\frac{{{5}{n}-{10}}}{{{3}{\left({3}{{n}}^{{2}}+{2}{n}-{4}\right)}}}<\frac{{{5}{n}}}{{{3}{\left({3}{{n}}^{{2}}-{4}\right)}}}$.

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

So, $\frac{{{5}{n}}}{{{3}{\left({3}{{n}}^{{2}}-{4}\right)}}}<\frac{{{5}{n}}}{{{3}\cdot{2}{{n}}^{{2}}}}=\frac{{5}}{{{6}{n}}}<\frac{{1}}{{n}}$.

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

Thus, ${x}_{{n}}\to\frac{{1}}{{3}}$.

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

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

This means that ${\left|{x}_{{n}}-{1}\right|}={\sqrt[{{n}}]{{{a}}}}-{1}<\frac{{{a}-{1}}}{{n}}<\epsilon$ when ${n}>{N}_{\epsilon}={\left[\frac{{{a}-{1}}}{\epsilon}\right]}$.

However, we can go another way.

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

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

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

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

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

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

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

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

Example 7. Now, consider infinite decreasing geometric sequence ${a},{a}{q},{a}{{q}}^{{2}},\ldots,{a}{{q}}^{{{n}-{1}}},\ldots$

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

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

We have that ${s}_{{n}}=\frac{{{a}-{a}{{q}}^{{n}}}}{{{1}-{q}}}=\frac{{a}}{{{1}-{q}}}-\frac{{a}}{{{1}-{q}}}{{q}}^{{n}}$.

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

Therefore, when ${\left|{q}\right|}<{1}$ we have that ${a}+{a}{q}+{a}{{q}}^{{2}}+\ldots+{a}{{q}}^{{{n}-{1}}}+\ldots=\frac{{a}}{{{1}-{q}}}$.