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How do I find this limit $ \lim \limits_{n \to \infty}(\dfrac{1}{n+1}+\dfrac{1}{n+2}.....\dfrac{1}{n+n})$

$(A)=\log_e1$

$(B)=\log_e2$

$(C)=\log_e3$

$(D)=\log_e4$

Please give me hint or something. What i could see is $ \lim \limits_{n \to \infty}\dfrac{1}{n} \rightarrow 0 $ that is not even in the options.

Damandeep
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  • While it's not the correct answer, $\ln 1 = 0$, so it was in the options. – orlp Feb 05 '18 at 12:09
  • Also: https://math.stackexchange.com/questions/1196920/is-the-following-statement-is-true, https://math.stackexchange.com/questions/1670508/solve-lim-n-to-infty-left-frac1n1-frac1n2-cdots-frac1nn, https://math.stackexchange.com/questions/73550/the-limit-of-truncated-sums-of-harmonic-series-lim-limits-k-to-infty-sum-n – Martin R Feb 05 '18 at 12:11
  • Am i going to jail? – Damandeep Feb 05 '18 at 12:13

2 Answers2

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$\dfrac{1}{n+1}+\dfrac{1}{n+2}.....\dfrac{1}{n+n}= \frac{1}{n}(\frac{1}{1+1/n}+\frac{1}{1+2/n}+....+\frac{1}{1+n/n}) \to \int_0^1 \frac{1}{1+x} dx= \log_e 2$

Fred
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Your reasoning is invalid, since the number of terms in the limit increases without bound. Regarding a solution, recall that $$\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n} \approx \log_\mathrm{e} n+\gamma , \quad n \to +\infty,$$ where $\gamma$ denotes the Euler-Mascheroni constant.

user1337
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