Lets discuss sequence, convergent sequence, divergent sequence and problems on them....

in #steemstem6 years ago

Sequence ,convergent sequence and divergent sequence


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Sequence


In simple words, sequence is enumerated collection of objects in which repetitions are allowed.

In field of real analysis , a sequence is defined as :-

If N is a set of natural number and X be any set,then a function f: N→X is called a sequence.
If X=R or subset of R (where R is set of real number), then f is called a real sequence and if X=C or subset of C (where C is a set of complex number) then f is called a complex sequence.

A sequence generally denoted by { }, by putting values in { }.

Examples of sequence:-

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Convergent Sequence


Convergent is an important property of sequence.A sequence is converge to a particular value known as limit of the sequence.

In field of real analysis, the convergent sequence is defined as:-

A sequence is said to converge to limit l , if given ε>0 , however small, there exists a positive integer m (depending upon ε) such that:-

|an -l |<ε for all n≥m

Problem on Convergent Sequence

Problem:-1) Show that the sequence {n+1/n} converges to 1.

Solution:- Let {an}={n+1/n}

Now consider given ε>0, however small, we choose a natural number m such that:-

m>1/ε or 1/m <1/ε

Therefore, for all n≥m or 1/n ≤1/m we have:-

|an -1| = |n+1-n/n| = 1/n but 1/n≤1/m <ε

Therefore, |an -1|<ε for all n≥m

⇒ limit n→∞an =1

Hence sequence {n+1/n} converges to limit 1.


Divergent Sequence


A sequence is said to be divergent, if it is not convergent.This means a divergent sequence doesn't converge to a finite value of limit. The value of limit of divergent sequence is infinite (+∞ to -∞)

In field of real analysis:-

  • A sequence {an} is said diverge to +∞ if k>0 however large, there exists a positive integer m(depending upon k) such that :-

a n > k for all n ≥ m

  • A sequence {an} is said diverge to -∞ if given k>0, however large, there exists a positive integer m(depending upon k) such that :-

a n <-k for all n ≥ m

Problem on divergent sequence

Problem:-2) Show that the sequence {n2 + 3n} diverges to +∞ .

Solution:- Let given sequence is {an} where an = n2 + 3n

Let k>0 however large--

Now an > k

if n2 + 3n > k

if n2 > k-3n

if n > √k - 3n

Now let a positive integer m just greater than √k - 3n

an > k for all n≥m

Therefore {n2 + 3n} diverges to +∞ .


Citations:-


Thanks for reading!


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