Theorems on Convergent Sequences with Proofs and Examples

The lecture covers theorems on convergent sequences, including the convergence of monotonic increasing and decreasing sequences when bounded. Detailed proofs for these theorems are provided, along with examples to determine if a sequence is bounded. The presentation includes step-by-step explanations and visual aids.

• Convergent sequences
• Monotonic sequences
• Sequence bounds
• Proof examples
• Mathematics

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1. Real Analysis Lecture-6 Theorems on Convergent Sequence Dated:-20.05.2020 PPT-18 UG (B.Sc., Part-2) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M.L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA

2. Theom3:-Every monotonic increasing sequence which is bounded above is convergent. i.e. converges to its least upper bound. na Proof:- Let be a monotonic increasing sequence (m.i.s) and bounded above. Since is m.i.s. hence . i e n na 1, a a n N + n n , m a a n m n a Again since is bounded above, So has upper bound i.e. it has some least upper bounds say k n a 0 , Th en for gi ven ( ) ( ii , i a k n N n ) , a k for a t least one va lue of n n Let the axiom (ii) be hold true for n=m Then m a , k

3. Proof continue But from (1), , , , n m a a So n m a a k n m n m , ....(2) k n N we get n m a k n m n (1), + , From axiom a n N n , a a k k k N n + , .....(3) n n (2) a (3), + from k and ( ) , k m n , ( ).....(1) a k n n = lim n a k n , . So a converges to k n Hence every monotonic increasing sequence which is bounded above is convergent.

4. Theom4:-Every monotonic decreasing sequence which is bounded below is convergent. i.e. converges to its greatest lower bound. na Proof:- Let be a monotonic decreasing sequence (m.d.s) and bounded below. Since is m.d.s. hence . i e n a 1, a a n n N + n n , ....(1) m a a n m n a Again since is bounded below, So has greatest lower bound i.e. it has some greatest lower bounds say M Then for giv en + n a 0, ( ) ( ) ii a , i a M M n N n , . for a t least one v alue of n n Let the axiom (ii) be hold true for n=m Then + , a M m

5. Proof continue But from (1), + + , , , n m a a So n n m a a M n m n m , ....(2) m a n N w n m a M n (1), , From the axiom a M a M from M N n N n M n , .... .(3) n n (2) (3 M ) + , and e ge t ( ) , M a m n ( ).....(1) , a M n n = lim n a M n , . So a conv e g r es to k n Hence every monotonic increasing sequence which is bounded above is convergent.

6. Examples Determine if the the sequence is bounded or not. + + + + 3 1 1 2 3! 1! n n n n ( ) i ( ) ii