Question #159884

Prove that a non-decreasing (resp. non-increasing) sequence which is not bounded above (resp. bounded below) diverges to +∞ (resp. to −∞).


1
Expert's answer
2021-02-03T15:17:31-0500

1) Let {an}n=1+\{a_n\}_{n=1}^{+\infty}  is a non-decreasing sequence which is not bounded above, that is, supnan=+\sup\limits_{n}a_n = +\infty. This means that for all M>0 there exists NNN\in\mathbb{N} such that aN>Ma_N>M. As the sequence {an}n=1+\{a_n\}_{n=1}^{+\infty} is a non-decreasing we have an>Ma_n>M for all n>N.n>N.

Finally: for all M>0 there exists NNN\in\mathbb{N} such that for all n>Nn>N an>Ma_n>M. Therefore, {an}n=1+\{a_n\}_{n=1}^{+\infty}diverges to +∞ by the definition.


2) Let {an}n=1+\{a_n\}_{n=1}^{+\infty} is a non-inreasing sequence which is not bounded below, that is, infnan=\inf\limits_{n}a_n = -\infty. This means that for all M<0 there exists NNN\in\mathbb{N} such that aN<Ma_N<M. As the sequence {an}n=1+\{a_n\}_{n=1}^{+\infty} is a non-increasing we have an<Ma_n<M for all n>N.n>N.

Finally: for all M<0 there exists NNN\in\mathbb{N} such that for all n>Nn>N an<Ma_n<M. Therefore, {an}n=1+\{a_n\}_{n=1}^{+\infty}diverges to -∞ by the definition.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS