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)

Expert's answer

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 < (L^{x}+ ^{1/2})^{1/x}

and

L > (L^{x}- ^{1/2})^{1/x}

there exists some N such that n ≥ N

We have a_{n}^{x}< L^{x}+

and

a_{n}^{x}> L^{x}-

This shows limn→∞ 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

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