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First of all, I know that Math Stack Exchange contains the answer but as I searched all the answer I can say that nobody used the simple limit rules to evaluate the limit. Either do I.

So here's my (minor) solution:

Here is $b=2, \quad a=1$, so $\Delta x = \dfrac{1}{n}$ and also there exist a point $x_{i}^{*}=1+\dfrac{i}{n}$. Let $f(x)=\dfrac{1}{x_{i}^{*}}$. Using Riemann Sum:

$$\displaystyle\lim_{n \to \infty} \displaystyle\sum_{i=1}^{n}= \dfrac{1}{n+i}$$

And more explicitly,

$$\displaystyle\lim_{n \to \infty} (\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}) $$

Here I can't solve the limit using just limit properties and some identities. Thanks.

Here's Answer!

beingmathematician
  • 687

1 Answers1

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Here is a way using that $\lim_{n\to\infty}(H_n-\log n)=\gamma $, where $\gamma $ is the Euler-Mascheroni constant and $H_n$ is the $n$-th harmonic number. Then note that

$$ \begin{align*} &\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}=H_{2n}-H_n\\[1em] &\therefore\quad 0=\lim_{n\to\infty}(H_{2n}-\log(2n)+\log n-H_n)=\lim_{n\to\infty}(H_{2n}-H_n)+\lim_{n\to\infty}(\log n-\log(2n))\\[1em] &\iff \lim_{n\to\infty}(H_{2n}-H_n)=\lim_{n\to\infty}(\log (2n)-\log n))=\lim_{n\to\infty}\log 2=\log 2 \end{align*} $$

Masacroso
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