Let $H_n=\sum_{k=1}^n{\frac{1}{k}}$

We investigate $H_{2^k}$

$H_{2^0}=H_1=\sum_{k=1}^1{\frac{1}{k}}=\frac{1}{1}=1+0(\frac{1}{2})$

$H_{2^1}=H_2=\sum_{k=1}^2{\frac{1}{k}}=\frac{1}{1}+\frac{1}{2}=1+1(\frac{1}{2})$

$\begin{align*}

H_{2^2}

&=H_4\\

&=\sum_{k=1}^4{\frac{1}{k}}\\

&=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\

&\gt \frac{1}{1}+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})\\

&\gt \frac{1}{1}+\frac{1}{2}+\frac{1}{2}\\

&\gt1+2(\frac{1}{2})

\end{align*}$

$\begin{align*}

H_{2^3}

&=H_8\\

&=\sum_{k=1}^8{\frac{1}{k}}\\

&=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\\

&\gt\frac{1}{1}+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})\\

&\gt \frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\\

&\gt1+3(\frac{1}{2})

\end{align*}$

By induction it can be shown that

$H_{2^k}\gt1+k(\frac{1}{2})$

Since we can make $k(\frac{1}{2})$ as large as we want by choosing a value for $k$ this implies that we can also make $H_{2^k}$ as large as want, i.e. $H$ diverges.

Started on MT365-TMA01/CMA41. Started on question 1 and got lost in the mechanics of GraphPlot and it's options. Learned some Mathematica again. I have to increase my efforts on MT365. Yellow Alert.

2-2018 Teaching by misleading

2 months ago

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