Could anyone help me solving the following limit?
$\lim_{n\rightarrow \infty} (\frac{1}{n+1}+ ... +\frac{1}{2n})$
I think it should just be 0 since we can distribute the limite inside the sum, but I am not sure of this answer. Thanks so much for your help!
We can approximate the sum by definite integrals from below and above in the following way (see here for more details) $$ \log\biggl(2-\frac1{n+1}\biggr)=\int_{n+1}^{2n+1}x^{-1}\mathrm dx\le\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}x^{-1}\mathrm dx=\log 2. $$ Since both of the bounds converge to the same limit as $n\to\infty$, we have that the limit as $n\to\infty$ is $\log 2$.
Notice $\sum_{i = 1}^n \frac{1}{n+i} = \sum_{i = 1}^{2n} \frac{1}{i} -\sum_{i = 1}^n \frac{1}{i} = H_{2n} - H_{n} $
Where $H_n$ stands for the $n^{th}$ term of harmoic numbers.
Therefore the limit equals $\lim_{n \to \infty } H_{2n} - H_{n}$
By the asymptotic expansion of $H_{n} ~ \ln(n) + \gamma + O(\frac{1}{n})$ when n goes up to infinity.
$$\lim_{n \to \infty } H_{2n} - H_{n} = \ln(2n) - \ln(n) = ln(2)$$
