I can't find the limits of:



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.

  • $\begingroup$ Instead of $x\mapsto\frac1{1+x}$ on $[0,1]$, why not consider $x\mapsto\frac1x$ on $[1,2]$? $\endgroup$ – Math1000 Nov 2 '17 at 10:52
  • $\begingroup$ HINT : 1) $H_{n} = \sum \frac{1}{n}$, then $H_{2n} - H_{n-1} = \sum_{k=n}^{2n} \frac{1}{k}$ $\endgroup$ – openspace Nov 2 '17 at 11:02
  • $\begingroup$ I'm a little confused by the expression for $b_n$, is the pattern for the exponent inside the radical supposed to be $n+4, n+8, n+12, \ldots, n+2n$ (implying $n$ is even I think?) or is it supposed to be $n+4, n+6, \ldots, n+2n$? $\endgroup$ – Rolf Hoyer Nov 2 '17 at 14:09
  • $\begingroup$ sorry I edited this $\endgroup$ – joe Nov 2 '17 at 14:31

$1)$ $$a_n=\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}=$$

$$=\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^n \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}}\right)$$

$$\lim_{n\to +\infty} a_n=\int_0^1 \frac{1}{1+x}\, dx=$$

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 $u=-2x$, $du=-2dx$.

$$=-\frac{1}{e}\int_0^{-2} e^u\, du=-\frac{1}{e}\left(e^u\right)\bigg|_{0}^{-2}=$$



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