Is this limit related to the harmonic series?

Calculus Level 3

A = lim ⁡ n → ∞ ( 1 n + 1 + 1 n + 2 + ⋯ + 1 2 n ) A=\lim_{n\to\infty}{\left(\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n} \right)}

Find the exact value of e A e^A if the limit exists, otherwise enter 0 as your answer.


The answer is 2.

Arturo Presa by Arturo Presa , 62, USA

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3 solutions

Rishabh Jain
May 3, 2016

I would definitely post a solution using Reimann Sums . Write the expression as: lim ⁡ n → ∞ 1 n ∑ r = 1 n ( 1 1 + r n ) = ∫ 0 1 d x 1 + x = ∣ ln ⁡ ( 1 + x ) ∣ 0 1 = ln ⁡ 2 \displaystyle\lim_{n\to\infty}\frac 1n\sum_{r=1}^n\left(\dfrac{1}{1+\frac rn}\right)=\displaystyle\int_0^1\dfrac{\mathrm{d}x}{1+x}=\left|\ln(1+x)\right|_0^1=\ln 2

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Otto Bretscher
May 3, 2016

It's good to see you again, compañero! Your problems are amongst the most thoughtful on Brilliant.

We have lim ⁡ n → ∞ ( H n − ln ⁡ ( n ) − γ ) = 0 \lim_{n\to \infty}(H_n-\ln(n)-\gamma)=0 so lim ⁡ n → ∞ ( ( H 2 n − ln ⁡ ( 2 n ) − γ ) − ( H n − ln ⁡ ( n ) − γ ) ) = 0 \lim_{n\to \infty}((H_{2n}-\ln(2n)-\gamma)-(H_n-\ln(n)-\gamma))=0 and A = lim ⁡ n → ∞ ( H 2 n − H n ) = ln ⁡ ( 2 ) A=\lim_{n\to\infty}(H_{2n}-H_n)=\ln(2) . The answer is 2 \boxed{2} .

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Thank you, compañero Otto! It is good to see you too. I think I have seen this problem already posted here before, but I am not sure. Thank you for your nice solution!

Arturo Presa - 5 years, 1 month ago
Arturo Presa
May 3, 2016

∑ k = n + 1 2 n 1 k = ∑ k = 1 2 n 1 k − 2 ∑ k = 1 n 1 2 k = ∑ k = 1 2 n ( − 1 ) k − 1 k , \sum_{k=n+1}^{2n} \frac{1}{k}\; = \; \sum_{k=1}^{2n} \frac{1}{k} - 2\sum_{k=1}^n \frac{1}{2k} \; =\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}, whose limit is ln ⁡ 2 \ln{2} as n n tends to ∞ . \infty. Therefore, e A = 2. e^A=2.

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