Question #64530

Q: Determine the following limits:
(a) lim((3√n)1/2n), (b) lim((n+1)1/ln(n+1)).

Expert's answer

Answer on Question #64530 – Math – Real Analysis

Question

Determine the following limits:

(a) lim((3νn)1/2n)\lim((3\nu n)1/2n), (b) lim((n+1)1/ln(n+1))\lim((n+1)1/ln(n+1))

Solution

(a) limn3n2n=32limn1n=0\lim_{n\to \infty}\frac{3\sqrt{n}}{2n} = \frac{3}{2}\lim_{n\to \infty}\frac{1}{\sqrt{n}} = 0

(b) We use the Stolz-Cesàro theorem:


limnanbn=limnanan1bnbn1,\lim _ {n \to \infty} \frac {a _ {n}}{b _ {n}} = \lim _ {n \to \infty} \frac {a _ {n} - a _ {n - 1}}{b _ {n} - b _ {n - 1}},


where an=n+1a_{n} = n + 1, the sequence bn=ln(n+1)b_{n} = \ln (n + 1) is strictly monotone (strictly increasing) and divergent (approaches ++\infty).

So we have


limnn+1ln(n+1)=limn(n+1)nln(n+1)lnn=limn1ln(n+1n)=1ln(limn(1+1n))=1/ln1=.\lim _ {n \to \infty} \frac {n + 1}{\ln (n + 1)} = \lim _ {n \to \infty} \frac {(n + 1) - n}{\ln (n + 1) - \ln n} = \lim _ {n \to \infty} \frac {1}{\ln \left(\frac {n + 1}{n}\right)} = \frac {1}{\ln \left(\lim _ {n \to \infty} \left(1 + \frac {1}{n}\right)\right)} = 1 / \ln 1 = \infty .


Answer: (a) 0; (b) \infty.

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