Answer on Question #64530 – Math – Real Analysis
Question
Determine the following limits:
(a) lim ( ( 3 ν n ) 1 / 2 n ) \lim((3\nu n)1/2n) lim (( 3 ν n ) 1/2 n ) , (b) lim ( ( n + 1 ) 1 / l n ( n + 1 ) ) \lim((n+1)1/ln(n+1)) lim (( n + 1 ) 1/ l n ( n + 1 ))
Solution
(a) lim n → ∞ 3 n 2 n = 3 2 lim n → ∞ 1 n = 0 \lim_{n\to \infty}\frac{3\sqrt{n}}{2n} = \frac{3}{2}\lim_{n\to \infty}\frac{1}{\sqrt{n}} = 0 lim n → ∞ 2 n 3 n = 2 3 lim n → ∞ n 1 = 0
(b) We use the Stolz-Cesàro theorem:
lim n → ∞ a n b n = lim n → ∞ a n − a n − 1 b n − b n − 1 , \lim _ {n \to \infty} \frac {a _ {n}}{b _ {n}} = \lim _ {n \to \infty} \frac {a _ {n} - a _ {n - 1}}{b _ {n} - b _ {n - 1}}, n → ∞ lim b n a n = n → ∞ lim b n − b n − 1 a n − a n − 1 ,
where a n = n + 1 a_{n} = n + 1 a n = n + 1 , the sequence b n = ln ( n + 1 ) b_{n} = \ln (n + 1) b n = ln ( n + 1 ) is strictly monotone (strictly increasing) and divergent (approaches + ∞ +\infty + ∞ ).
So we have
lim n → ∞ n + 1 ln ( n + 1 ) = lim n → ∞ ( n + 1 ) − n ln ( n + 1 ) − ln n = lim n → ∞ 1 ln ( n + 1 n ) = 1 ln ( lim n → ∞ ( 1 + 1 n ) ) = 1 / ln 1 = ∞ . \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 . n → ∞ lim ln ( n + 1 ) n + 1 = n → ∞ lim ln ( n + 1 ) − ln n ( n + 1 ) − n = n → ∞ lim ln ( n n + 1 ) 1 = ln ( lim n → ∞ ( 1 + n 1 ) ) 1 = 1/ ln 1 = ∞.
Answer: (a) 0; (b) ∞ \infty ∞ .
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