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Evaluate $$\lim\Big(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\Big)$$ using sequential methods.

Of course: $$\frac{1}{n}\bigg(\frac{1}{1+1/n}+\frac{1}{1+2/n}+\ldots+\frac{1}{1+n/n}\bigg) \rightarrow \int_{0}^{1}\frac{dx}{1+x}=\ln 2$$

but by using only sequences, I don't know where to start from. I thought of the squeeze theorem, but i am not sure how to get $\ln 2$ on the way.

Thanks in advance.

Nikolaos Skout
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    I've just added an alternative by a mixed method (integral bound + squeeze theorem), I hope you can find also that way useful as an alternative method to those already presented here. Bye – user Sep 28 '18 at 10:12
  • In this answer, the limit is derived using only Bernoulli's Inequality, Telescoping Products, and the Squeeze Theorem (no integrals). – robjohn Mar 15 '23 at 16:08

5 Answers5

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Well , it is quite well known that $ H_n - \log(n) \to \gamma $ ( where $ H_n= \sum_{k=1}^{n} \frac{1}{k} $ and $ \gamma $ is the euler-mascheroni constant )

Denote by $ a_n $ your sequence , we have that $$ a_n= H_{2n}-H_{n}=H_{2n}-\log(2n)+\log(2n)-H_n=H_{2n}-\log(2n)+\log(2)+\log(n)-H_n.$$

Hence $\lim_{n \to \infty} a_n = \gamma+\log(2)-\gamma=\log(2).$

Michael Hardy
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Impending Uncertainty
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Using the fact that $\ln(1 + x) \leq x$, we have $$ \ln(k + 1) - \ln k \leq \frac{1}{k} \leq \ln{k} - \ln(k - 1), $$ hence $$ \ln \frac{2n + 1}{n + 1} \leq \sum_{k = {n + 1}}^{2n} \frac{1}{k} \leq \ln \frac{2n}{n} = \ln2, $$ and the squeeze theorem will give the result.

Xiangxiang Xu
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We have

$$\sum_{k=1}^n \frac1{n+k}=\frac1n\sum_{k=1}^n \frac1{1+k/n}=\frac1n\sum_{k=1}^n\sum_{j=0}^\infty (-1)^j\left(\frac k n\right)^j=\ldots$$

and for any fixed $n$ since by Faulhaber's formula

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

we have

$$\ldots =\sum_{j=0}^\infty \frac{(-1)^j}{n^{j+1}}\sum_{k=1}^nk^j = \sum_{j=0}^\infty \frac{(-1)^j}{j+1}+O\left(\frac1n\right)\sum_{j=0}^\infty (-1)^j$$ $$\to\sum_{j=0}^\infty \frac{(-1)^j}{j+1}=1-\frac12+\frac13-\frac14+\ldots=\ln 2$$

indeed by Alternating harmonic series the result follows.

user
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  • This asserts with no justification that, if, for every $j$, $x_n(j)\to x(j)$ when $n\to\infty$, then $\sum\limits_{j=1}^\infty x_n(j)\to\sum\limits_{j=1}^\infty x(j)$ when $n\to\infty$. As everybody should know, this does not hold in general (in fact, asserting that it does is a cool way to "prove" that $1=0$...). Presenting a proof that it does in the present case would be starting to actually do some mathematics. – Did Sep 28 '18 at 10:12
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As an alternative by integral bounds since

$$\ln \left(\frac{2n+1}{n+1}\right)=\int_{n+1}^{2n+1}\frac1x dx \le \sum_{k=1}^{n} \frac1{n+k} \le \frac1{n+1}+\int_{n+2}^{2n+1}\frac1x dx=\frac1{n+1}+\ln \left(\frac{2n+1}{n+2}\right)$$

by squeeze theorem the result follows.

user
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We can write sum as

$$\sum^{n}_{k=1}\frac{1}{n+k}=\sum^{2n}_{k=1}\frac{1}{k}-2\sum^{n}_{k=1}\frac{1}{2k}$$

So $$\sum^{n}_{k=1}\frac{1}{n+k}=1-\frac{1}{2}+\frac{1}{3}+\cdots\cdots +\frac{1}{2n-1}-\frac{1}{2n}$$

Now $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n+k}=1-\frac{1}{2}+\frac{1}{3}-\cdots\cdots =\ln(2).$$

DXT
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