Real Analysis Lecture Problems and Solutions
Explore the solutions to convergent sequences and binomial theorems in Real Analysis lecture problems. Understand the proofs step by step for various mathematical concepts.
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Real Analysis Lecture-9 Dated:-27.05.2020 PPT-22 UG (B.Sc., Part-2) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M.L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA
Problems 1 1. Prove that 2. Show that the sequence , where 1 1 1 1 2 3 n n n + + + = = n 1 . . lim n n a 1. n as n ie n n 1 n + l is convergent and = + + + a n n 1 2 its limit is l . where 3. Show that the sequence is convergent and 4. Prove that ( ( n 1. ( ) n 1 = + 1 a n n ( ) n 1 + = lim 1 n , 2 3. e where e n ) ( ) + 1 n n 1 1 + = + = ( ) lim 1 n ( )lim 1 n i e ii e + 1 n n ) n 2 + = 2 ( )lim 1 iii e n
Solution of (1) Let , then we have to prove that Let where L is very small positive quantity. 1 , L = + = lim n 1 a 1n = a n n n na 1 = + 1 n L n ( 1) ( 1)( 3! 2) n n n n n = + = + + + + 2 3 n (1 ) 1 n L nL L L 2! ( 1) 2 n n n 2 2 n L L 2! 2 n ( 1) 2 n n 2 L L ( 1) ( 1) n 2 lim n lim n lim n 0 ( ) L L negative ( 1) n . But L is a very small positive quantity So L = lim n 0 ( ) = + = + = lim n lim 1 n = 1 0 1 a L n 1 lim n 1 n proved n
Solution of (2) Given 1 + 1 + 1 + 1 + = + + + , a then n 1 1 + 2 1 + 3 1 + n n n n n 1 + = + + + a + 1 n 2 3 4 2 2 1 n n n n n 1 + 1 + 1 + 1 2 1 n 1 + = + + + + + + + 2 3 4 2 1 2 2 n n n n n 1 1 + = + , Now a a + 1 n n + + 2 1 2 2 1 n n 1 1 1 = = 0, n N + + + + 2 1 2( 1) (2 N 1)(2 2) n a a n n n 0, n a n N + 1 n n , a + 1 n n So, is monotonic increasing sequence. na
Continue Again 1 + 1 + 1 + 1 + = + + + a n 1 2 3 n n n n n 1 n 1 n 1 n 1 n 1 n + + + = = 1 n 1, ....(1) a n N n Also 1 + 1 + 1 + 1 + = + + + a n 1 2 3 n n n n n 1 + 1 + 1 + 1 + + + + n n n n n n n n 1 2 1 2 1 2 1 2 1 2 1 2 = + + + = = n n n n n n 1, 2 ....(2 ) a n N n 1 2 From (1) and (2), we get, Thus is monotonic increasing and bounded. So, is convergent and has unique limit say l . Hence 1 1, 2 1, n a n N n a n a l proved
Solution of (3) ( ) Given n 1 = + 1 ....(1) a n n By binomial theorem 2 3 1 n ( 1) 1 n ( 1)( 3! 2 n 2) 1 n n n n n n n = + + + + 1 a n 1! 1 2! 1 2! 2! 1 n 1 3! 1 n 1 1 = + + + + 1 1 1 1 3! 1 4! 1 n 1 1 + + + + + + ! ( ) n 1 1 1 2 1 2 1 2 1 2 1 n 1 1 + + + + + + = + 1 1 2 3 1 2 1 2 1 2 1 n = + = + 1 2 1 1 2 3 1 n 2 3, ... (2) . a n N n
Continue Also + e where 1 2! 1 n 1 3! 1 n = 2 n 1 n 1 1 = + + + + + 1 1 1 a n ! ( ) n 1 1 2! 1 n 1 3! 1 n 2 n lim 1 n 2 3 e n = + + + 2 1 1 1 1 n 2 n 3 n 2, ..( 1 0,1 0, 1 0, ) a n N as n . . 2 ie , ....(3) a n N n From (2) and (3), We get, Therefore, the sequence is bounded. Now, Thus is monotonic increasing and bounded sequence and hence it is convergent. And 2 3, n a n N n a 0, 0, a a a a n n N N + 1 n n + 1 n n n a
