How to show the limit is $\log(2)$? 
Show that limit of string $\left(a_{n}\right)_{n\ \geq\ 1}$ is $\log\left(2\right)$, where
$$
a_{n} = \int_{1/\left(2n\right)}^{1/n}
{\mathrm{e}^x \over x}\,\mathrm{d}x
$$.

My approach: It's easy to show that using
$\mathrm{e}^{x} \geq x + 1$ and you got right side.
But what about left side ?. Any ideas ?. $\ldots$
 A: HINT:
METHODOLOGY $1$:  Bounding the Exponential Function
If we use the inequalities $1+x\le e^x\le \frac1{1-x}$, for $x<1$, we have for $n>1$
$$\int_{1/(2n)}^{1/n}\frac{1+x}{x}\,dx\le \int_{1/(2n)}^{1/n}\frac{e^x}{x}\,dx\le \int_{1/(2n)}^{1/n}\frac{1}{x(1-x)}\,dx$$

HINT:
METHODOLOGY $2$:  Integrating by Parts
Using integration by parts, note that we have
$$\int_{1/(2n)}^{1/n} \frac{e^x}{x}\,dx=\log(1/n)e^{1/n}-\log(1/(2n))e^{1/(2n)}-\int_{1/(2n)}^{1/n}\log(x)e^x\,dx$$
Now show that 
$$\lim_{n\to \infty}\left(\log(1/n)e^{1/n}-\log(1/(2n))e^{1/(2n)}\right)=\log(2)$$
and 
$$\lim_{n\to\infty}\int_{1/(2n)}^{1/n}\log(x)e^x\,dx=0$$

A: Hint:
$e^{1/2n}(1/x) < e^x/x <e^{1/n}(1/x);$
$e^{1/2n}\displaystyle{\int_{1/2n}^{1/n}}(1/x) dx < \int_{1/2n}^{1/n}(e^x/x) dx < e^{1/n}\int_{1/2n}^{1/n}(1/x) dx$
Used: Monotony of Riemann integrals.
A: Hint:
Using $\exp(x)=\sum_{k=0}^{\infty}\frac{x^k}{k!}$, you will get $$
\lim_{n \to \infty}\int_{1/n}^{1/2n} \left( \frac{1}{x}+1+\cdots \right) dx=\log 2$$
A: Let $ n $ be a positive integer different from $ 0 $ :
Notice that $ x\mapsto\frac{\mathrm{e}^{x}-1}{x} $ is continuous on $ \left(0,+\infty\right) $, and can be extended to a continuous function on $ \left[0,+\infty\right) $, which is the reason why it is integrable on any segment $ \left[0,a\right] $, where $ a\geq 0 $.
\begin{aligned}\int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}}{x}\,\mathrm{d}x}&=\int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}+\ln{2}\\ &=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}-\int_{0}^{\frac{1}{2n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}+\ln{2}\\ &=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-1}{x}\,\mathrm{d}x}-\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{\frac{y}{2}}-1}{y}\,\mathrm{d}y}\\ \int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}}{x}\,\mathrm{d}x}&=\int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x}\,\mathrm{d}x}+\ln{2}\end{aligned}
Notice that in the third line we substituted $ x=\frac{y}{2}\cdot $
Since $ x\mapsto\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x} $ is also a continuous function on $ \left(0,+\infty\right) $, and can be extended to a continuous function on $ \left[0,+\infty\right) $, it can be upper-bounded on $ \left[0,1\right] $ by some constant $ M\geq 0 \cdot $
Thus : $$ \int_{0}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}-\mathrm{e}^{\frac{x}{2}}}{x}\,\mathrm{d}x}\leq\frac{M}{n}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence : $$ \lim_{n\to +\infty}{\int_{\frac{1}{2n}}^{\frac{1}{n}}{\frac{\mathrm{e}^{x}}{x}\,\mathrm{d}x}}=\ln{2} $$
A: In fact you can integrate Taylor expansions, more precisely 
if $f(x)=a_0+a_1x+\cdots+a_nx^n+o(x^n)$ 
then $F(x)=F(0)+a_0X+\cdots+\frac{a_n}{n+1}x^{n+1}+o(x^{n+1})$
Using $e^x=1+x+o(x)$ we get $\frac{e^x}{x}=\frac 1x+1+o(1)$ with a term $\frac 1x$ not verifying the conditions.
But simply set $f(x)=\frac{e^x-1}{x}=1+o(1)$ and you can apply the theorem.
And get $\displaystyle\int_{1/2n}^{1/n}\frac{e^x}{x}\mathop{dx}=\int_{1/2n}^{1/n}\frac{1}{x}\mathop{dx}+\frac 1{2n}+o\left(\frac 1n\right)\to\ln 2$
The reason being that it's just inequalities in disguise (i.e. $|o(1)|<\varepsilon$ for $n\gg 1$).
