Question

Every convergent sequence is bounded what about is converse? Prove both statements with examples​

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It's true, that every convergent sequence is bounded, but is every bounded sequence convergent?How do we prove that every convergent sequence bounded?

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By convergent sequence, do you mean the convergent sequence of real numbers ? If yes, then you can proceed as follows:

Let (an)∞n=0 be a convergent sequence of real numbers. By definition of a convergent sequence, the sequence (an)∞n=0 converges to L for some L∈R.

Since (an)∞n=0 converges to L , for every ε>0 , we can find N∈N such that d(an,L)≤ε for every n≥N . In other words, we have:

∀ε>0∃N∈N|an−L|≤ε∀n≥N.

In particular, for ε=1 , we can find N∈N such that |an−L|≤1 for every n≥N. But |an−L|≤1 implies that L−1≤an≤L+1 , which in turn would imply that −|L|−1≤an≤|L|+1 . i.e., |an|≤|L|+1. We thus have |an|≤|L|+1 for every n≥N

Using induction, you can show that the finite sequence (an)N−1n=0 is bounded .i.e., there is M∈R>0 such that |an|≤M for all 0≤n≤N−1.

Now choose K:=max{M,|L|+1} . Then we will have |an|≤K for all n∈N , which shows that the sequence (an)∞n=0 is bounded.

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Take any (xn) with xn→x for some real number x . Then there must exist a natural number M such that for all m≥M , |xm−x|<1 , and hence |xm|<|x|+1 . But then |xm|≤max{|x|+1,|x1|,…,|xM|} for all m∈N . Hence, (xn) is bounded