TITLE: What would Gauss do in this case: adding $1+\frac12+\frac13+\frac14+ \dots +\frac1{100}$?
QUESTION [30 upvotes]: We all know the story related to Gauss that Gauss' class was asked to find the sum of the numbers from $1$ to $100$ as a "busy work" problem and and he came up with $5050$ in less than a minute. He used a simple trick  $50\times 101=5050$  there.
Now what if in some parallel universe, his teacher knew Gauss would figure that out quickly and  asked the class to calculate $$1+\frac12+\frac13+\frac14+ \dots +\frac1{100}$$ instead, and assures himself a nice nap. 
Is there any way Gauss could still impress the world in that universe by calculating it precisely up to, say, two decimal points using some trick (assuming he knows advanced mathematics too, although still in junior class). I do not see any quicker way to find this sum and had to use wolfram alpha which gives
$$\frac{14466636279520351160221518043104131447711}{2788815009188499086581352357412492142272}  \approx 5.1873.$$ 
What is the best method/trick to reach around $5.1$ or even $5$ quickly than any other student in your class, and impress the world?
We can make G.P's like $(1+\frac12+\frac14+\frac18+\frac1{16}+\frac1{32}+\frac1{64})+(\frac13+\frac19+\frac1{27}+\frac1{81})$ but we still leave way too many terms out of the G.P.'s and will have to find them separately by dividing.

REPLY [4 votes]: Even on an off day, young Gauss would have seen at a glance that
$$\begin{align}
1+{1\over2}+\cdots+{1\over10}&=\left(1+{1\over2}+{1\over3}+{1\over6}\right)+\left({1\over4}+{1\over5}+{1\over8}+{1\over10} \right)+\left({1\over7}+{1\over9} \right)\\
&=2+(.25+.2+.125+.1)+(.142857...+.111111...)\\
&=2.675+.253968...\\
&=2.928968...
\end{align}$$
A suitably precocious Gauss would have sensed that
$${1\over11}+\cdots+{1\over100}\approx\int_{10.5}^{100.5}{dx\over x}=\ln\left(201\over21 \right)=\ln\left(67\over7 \right)=\ln10+\ln\left(1-{3\over70} \right)$$
and he would have known that $\ln10=2.302585...$ and $\ln(1-{3\over70})\approx-{3\over70}=-.042857...$
Finally, young Gauss would have figured that a decent two-digit approximation would be
$$2.93+2.30-.04=5.19$$
What unclear (to me, at least) is how he would have handled the error estimates to know how close he'd come.