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how can I solve this?

$$\lim_{n\to \infty } \left(\frac{1}{1+n} + \frac{1}{2+n} + ... + \frac{1}{2n} \right) $$

but the limit of this?

$$(x_n) = \frac{1}{\ln(n^2)} + \frac{1}{\ln(n^3)} + ... + \frac{1}{\ln(n^n)} $$

At the second sequence I tried this:

$$ (x_n) = \frac{1}{\ln(n)}\Bigl(\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}\Bigr)$$ That is indetermination $ 0\cdot \infty$ since the sum diverges, but I don't know how to continue.

Bernard
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Loop24
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    You're asking two questions in one. We generally prefer question posts to focus on a single question. Also, what have you tried? What tools for limit calculations do you have at your disposal? What's stopping you from applying them? – Arthur Aug 09 '19 at 11:05
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    By the way, Approach0 is really useful to find old questions on this site. That's how I found the duplicate I mentioned in my first comment. – Arnaud D. Aug 09 '19 at 11:21

1 Answers1

1

We will use the following fact.

Lemma. There is a real number $\gamma$ such that $$ \lim\limits_{n\rightarrow\infty}\left(H_n-\ln n\right) = \gamma, $$ where $H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$ is a $n$-th harmonic number.

Now, it's clear that $$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}=H_{2n}-H_n. $$ Hence, $$ \lim\limits_{n\rightarrow\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\right)= \\ \lim\limits_{n\rightarrow\infty}\left(H_{2n}-H_n\right)= \lim\limits_{n\rightarrow\infty}(H_{2n}-\ln (2n))-\lim\limits_{n\rightarrow\infty}(H_n-\ln n) + \ln 2 = \gamma-\gamma+\ln 2 = \ln 2. $$

For the second one note that from the lemma we obtain that $$ \lim\limits_{n\rightarrow\infty}\frac{H_n-\ln n}{\ln n} = 0,~\text{so}~\lim\limits_{n\rightarrow\infty}\frac{H_n}{\ln n} = 1. $$ Thus, $$ \lim\limits_{n\rightarrow\infty}\frac{\frac{1}{2}+\ldots+\frac{1}{n}}{\ln n} = \lim\limits_{n\rightarrow\infty}\frac{H_n-1}{\ln n} = 1. $$

richrow
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  • Why do you do so complicated for such an easy exercice ? (-1) – Surb Aug 09 '19 at 11:24
  • Because I wanted to avoid using properties of integrals. This solution uses only sequence limits. The main lemma follows from convergence of decreasing and bounded sequence. – richrow Aug 09 '19 at 11:54