# Infinitely Small Sequence

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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, |x_n-0|=|x_n|<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 |alpha_n|=|x_n-a|<epsilon. And vice versa, if alpha_n is infinitesimal then x_n->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->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=1/n, x_n=-1/n, x_n=((-1)^(n+1))/n.

Corresponding lists are

{1,1/2,1/3,1/4,...},

{-1,-1/2,-1/3,-1/4,...},

{1,-1/2,1/3,-1/4,...}.

All variants are infinitesimal because |x_n-0|=|x_n|=|1/n|<epsilon when n>1/epsilon. Therefore, we can take (recall that N is natural number) N=N_epsilon>[1/epsilon], where [x] 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=(2+(-1)^n)/n.

Corresponding list is {1,3/2,1/3,3/4,1/5,1/2,...}.

By triangle inequality we have that |x_n|=|2/n+((-1)^n)/n|<=|2/n|+|((-1)^n)/n|=|2/n|+|1/n|=|3/n|.

Therefore, |3/n|<epsilon when n>3/epsilon . So, N_epsilon=[3/epsilon]. Thus, this variant is infinitesimal.

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

Example 3. Consider sequence x_n=(1+(-1)^n)/n.

Corresponding sequence is {0,1,0,1/2,...}.

By triangle inequality we have that |x_n|=|1/n+((-1)^n)/n|<=|1/n|+|((-1)^n)/n|=|1/n|+|1/n|=|2/n|.

Therefore, |2/n|<epsilon when n>2/epsilon . So, N_epsilon=[2/epsilon]. 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=(n^2-n+2)/(3n^2+2n-4). Show that its limit is 1/3.

To show that limit of this sequence is 1/3 we need to show that |x_n-1/3| is infinitesimal.

So, after algebraic manipulations we obtain that |x_n-1/3|=|(n^2-n+2)/(3n^2+2n-4)-1/3|=|-(5n-10)/(3(3n^2+2n-4))|=(5n-10)/(3(3n^2+2n-4))<(5n)/(3(3n^2-4)).

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

So, (5n)/(3(3n^2-4))<(5n)/(3*2n^2)=5/(6n)<1/n.

Therefore, |x_n-1/3|<epsilon when 1/n<epsilon or n>1/epsilon. In other words if n>N_epsilon=[1/epsilon] .

Thus, x_n->1/3.

Example 5. Let sequence is given by formula x_n=a^(1/n)=root(n)(a) (a>1). Let's prove that x_n->1.

Using Bernoulli inequality gamma^n>1+n(gamma-1) where n is natural and n>1, gamma>1, with gamma=a^(1/n), we obtain that (a^(1/n))^n>1+n(a^(1/n)-1) or a^(1/n)-1<(a-1)/n.

This means that |x_n-1|=root(n)(a)-1<(a-1)/n<epsilon when n>N_epsilon=[(a-1)/epsilon].

However, we can go another way.

Inequality |x_n-1|=a^(1/n)-1<epsilon is equivalent to the inequality 1/n<log_a(1+epsilon) or n>1/(log_a(1+epsilon)), so it holds when n>N_epsilon=[1/(log_a(1+epsilon))].

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)=9/(0.01)=900 according to the first way and N_(0.01)=[1/(0.00432....)]=231 according to the second way. Second way gave us the smallest of all possible values for N_(0.01) because 10^(1/231)=1.010017 is different from 1 on more than epsilon=0.01.

Same will be in general case, because when a<=1/(log_a(1+epsilon)) we have that a^(1/n)-1>=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>=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 |q|<1.

To prove that alpha_n->0 consider inequlaity |alpha_n|=|q|^n<epsilon. This inequality is equivalent to the inequality lg(|q|^n)<lg(epsilon) or nlg|q|<lg(epsilon) i.e. n>(lg(epsilon))/(lg|q|).

Therefore, if we take (considering epsilon<1) N_epsilon=[(lg(epsilon))/(lg|q|)] then for n>N_epsilon above inequality will hold.

Similarly it can be shown that variant beta_n=A*q^n, where |q|<1 and A is constant, is also infinitesimal.

Example 7. Now, consider infinite decreasing geometric sequence a,aq,aq^2,...,aq^(n-1),...

where |q|<1. Let's find its sum.

Sum of infinite progression is limit as n->oo of partial sum of geometric progression s_n.

We have that s_n=(a-aq^n)/(1-q)=a/(1-q)-a/(1-q)q^n.

This equality means that variant s_n is different from number a/(1-q) on alpha_n=1/(1-q) q^n. But as we saw in example 6 alpha_n is infinitesimal. Therefore, s=lim(s_n)=1/(1-q).

Therefore, when |q|<1 we have that a+aq+aq^2+...+aq^(n-1)+...=a/(1-q).