# Question #e5cde

Your sum can be expressed as a Riemann Sum. Hence

$\setminus \frac{1}{n + 1} + \ldots + \setminus \frac{1}{2 n} = \setminus {\int}_{n}^{2 n} \setminus \frac{1}{x} \mathrm{dx} = \ln \left(2 n\right) - \ln \left(n\right) = \ln \left(2\right) \approx 0.693$

Also notice that

$\frac{1}{2} = \setminus \frac{n}{2 n} \setminus \le \setminus \frac{1}{n + 1} + \setminus \frac{1}{n + 2} + \setminus \ldots + \setminus \frac{1}{2 n} \setminus \le \setminus \frac{n}{n + 1}$

As well as

$\setminus {\sum}_{k = 1}^{n} \setminus \frac{1}{n + k} \setminus \le q 1 - \setminus {\sum}_{k = 1}^{n} \setminus \frac{k}{2 {n}^{2}} = 1 - \setminus \frac{n \left(n + 1\right)}{4 {n}^{2}} = \frac{3}{4} - \setminus \frac{1}{4 n}$

because

$\setminus \frac{1}{n + k} \setminus \le q \setminus \frac{1}{n} - \setminus \frac{k}{2 {n}^{2}}$

Mar 30, 2016

The limit is $\ln 2$, but I don't see how to get it using the sandwich theorem.

#### Explanation:

We want ${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \frac{1}{n + i}$

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \frac{1}{1 + \left(\frac{i}{n}\right)} \frac{1}{n}$

This is an integral.

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$

Where $\Delta x = \frac{b - a}{n}$ $\text{ }$ and $\text{ }$ ${x}_{i} = a + i \Delta x$

Here we can have $\Delta x = \frac{1}{n}$
and use either

$a = 1$ so $b = 2$ and ${x}_{i} = 1 + \frac{i}{n}$
In which case $f \left(x\right) = \frac{1}{x}$ and we have ${\int}_{1}^{2} \frac{1}{x} \mathrm{dx} = \ln 2$

Or we can use

$a = 0$ so $b = 1$ and ${x}_{i} = \frac{i}{n}$
In which case $f \left(x\right) = \frac{1}{1 + x}$ and we have ${\int}_{0}^{1} \frac{1}{1 + x} \mathrm{dx} = \ln 2$