Per the title of this question, how does one go about calculatinglimn→∞1n2(ln⁡(2n3n)+ln⁡(5n4n)+⋯+ln⁡((3n−1)n(n+2)n))

The k-th term isnlog⁡(3k−1)k+2=nlog⁡3+5n3k+O(nk2)and so your sum is1n2∑k=1n(nlog⁡3+5n3k+O(nk2))=1n2(n2log⁡3+5nlog⁡n3+O(n))Here we used the estimate ∑k=1n1k=log⁡n+O(1)Thus the sum is=log⁡3+5log⁡n3n+O(1n)and so your limit islog⁡3

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High School
Calculus and Analysis
Calculus 1
Limits and continuity

Abbigail Robles

Answered question

2022-04-24

Per the title of this question, how does one go about calculating

lim_{n \to \infty} \frac{1}{n^{2}} (\ln (\frac{2^{n}}{3^{n}}) + \ln (\frac{5^{n}}{4^{n}}) + \dots + \ln (\frac{{(3 n - 1)}^{n}}{{(n + 2)}^{n}}))

Answer & Explanation

Brenton Steele

Beginner2022-04-25Added 18 answers

The k-th term is

n \log \frac{(3 k - 1)}{k + 2} = n \log 3 + \frac{5 n}{3 k} + O (\frac{n}{k^{2}})

and so your sum is

\frac{1}{n^{2}} \sum_{k = 1}^{n} (n \log 3 + \frac{5 n}{3 k} + O (\frac{n}{k^{2}}))

= \frac{1}{n^{2}} (n^{2} \log 3 + \frac{5 n \log n}{3} + O (n))

Here we used the estimate

\sum_{k = 1}^{n} \frac{1}{k} = \log n + O (1)

Thus the sum is

= \log 3 + \frac{5 \log n}{3 n} + O (\frac{1}{n})

and so your limit is

\log 3

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