I can't find the limits of:
1.$a_n=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$
2.$b_n=\frac{2}{n\sqrt[n]{e^{n+2}}}+\frac{2}{n\sqrt[n]{e^{n+4}}}+...+\frac{2}{n\sqrt[n]{e^{n+2n}}}$
I need to use Riemann sum in order to solve it.
Now, in limit 1 I found that if $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}$ then I can use Riemann sum of the function $\frac{1}{1+x}$ in $[0,1]$ but $a_n$ starts with $\frac{1}{n}$.
I have no idea how to solve limit 2.
Thanks for the help.
$1)$ $$a_n=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}=$$
$$=\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^n \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}}\right)$$
$$\lim_{n\to +\infty} a_n=\int_0^1 \frac{1}{1+x}\, dx=$$
Let $u=1+x$, $du=dx$. $$=\int_1^2 \frac{1}{u}\, du=(\ln|u|)\bigg|_1^2=$$
$$=\ln|2|-\ln|1|=\ln 2-0=\ln 2$$
$2)$ $$b_n=\frac{2}{n\sqrt[n]{e^{n+2}}}+\frac{2}{n\sqrt[n]{e^{n+4}}}+\cdots+\frac{2}{n\sqrt[n]{e^{n+2n}}}=$$
$$=\sum_{k=1}^{n}\frac{2}{n\sqrt[n]{e^{n+2k}}}=\sum_{k=1}^n \frac{2}{n\sqrt[n]{e^n}\sqrt[n]{e^{2k}}}=$$
$$=\sum_{k=1}^{n}\frac{2}{n\cdot e\sqrt[n]{e^{2k}}}=\frac{2}{e}\sum_{k=1}^{n}\frac{1}{n}\left(\frac{1}{\left(e^2\right)^{\frac{k}{n}}}\right)$$
$$\lim_{n\to +\infty} b_n=\frac{2}{e}\int_{0}^1 \frac{1}{e^{2x}}\, dx=$$
Let $u=-2x$, $du=-2dx$.
$$=-\frac{1}{e}\int_0^{-2} e^u\, du=-\frac{1}{e}\left(e^u\right)\bigg|_{0}^{-2}=$$
$$=-\frac{1}{e}(e^{-2}-e^0)=-\frac{1}{e^3}+\frac{1}{e}$$