Answer to Question #178664 in Real Analysis for Chanchal

Question #178664
  • Let(a↓n) ↓n€B be any sequence. Show that lim↓n-->♾️ a↓n =Liff for every E>0, there exists some N€N such that n>=N implies a↓n N↓e (L)
1
Expert's answer
2021-04-20T16:24:21-0400



Given

Let(a↓n) ↓n€B be any sequence. Show that lim↓n-->♾️ a↓n =Liff for every E>0, there exists some N€N such that n>=N implies a↓n N↓e (L)

By this we observed


To prove



Solution. Let > 0. By Theorem , note that

L < (Lx+ 1/2)1/x


and

L > (Lx- 1/2)1/x



there exists some N such that n ≥ N

We have anx< Lx+

and

anx> Lx-


This shows limn→∞ anx = Lx


.





Or


we can also solve by this way



Prove that {an} is a Cauchy sequence. Solution .

First we prove by induction on n that |an+1 − an| ≤ αn−1|a2 − a1| for all ...


an < A + ε. 2. ,

and there exists N2 such that for all n>N2,


bn < B + ε. 2 . But then if N = max(N1, N2),


then for any n>N we have an + bn < A + .

..xn − 1,n ∈ N.

Show that (xn) is decreasing and bounded below by 2. Find the ... 2 + xn,n ∈ N.

Show (xn) converges and find the limit. ...

2+2 = 4,


so xn < 2 for all n ∈ N by induction




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