Answer to Question #178664 in Real Analysis for Chanchal

Question

• 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)

Answer 1

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

$\bigstar$

To prove $\boxed{lim\space n→∞ \space a^n_x = L^x}$

Solution. Let $\epsilon$> 0. By Theorem , note that

$\bull$

L < (L^x+ $\epsilon$ ^1/2)^1/x

and

$\bull$

L > (L^x- $\epsilon$ ^1/2)^1/x

there exists some N such that n ≥ N

$\bull$

We have a_n^x< L^x+ $\epsilon$

and

a_n^x> L^x-$\epsilon$

$\bigstar$

This shows limn→∞ a_n^x = L^x

.$\boxed{lim\space n→∞ \space a^n_x = L^x}$

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