What would Gauss do in this case: adding $1+\frac12+\frac13+\frac14+ \dots +\frac1{100}$? 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.
 A: 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.
A: Euler's identity would be easiest:
$$H_n \approx \ln n + \gamma$$
Then (assuming no calculators) you would remember that $\ln 10 \approx 2.3$,  so that $\ln 100 \approx 4.6$, getting that:
$$H_{100} \approx 4.6+ 0.577 \approx 5.18$$
Now, one may argue that this is a bit of a cheat, since, $\gamma$ is sort of defined by the difference between $H_n$ and $\ln n$. What one could do however, is calculate $\gamma$ knowing that most of the contribution comes from the first few members of the series, since the difference between the sum and the integral becomes smaller as the derivative becomes smaller. This is something an older Gauss could have potentially worked out without Euler's work.
Let's take $n=10$ for example:
$$\gamma \approx H_{10} - \ln 10$$
So that we get:
$$H_{100} \approx \ln 10 + H_{10}\approx 2.3 + 2.9 = 5.2$$
Which is pretty close, considering all you had to do was sum up the first $10$ numbers, which only involves a single long division!
A: The approximation $$H_n\approx\log(2n+1)$$
https://math.stackexchange.com/a/1602945/134791
could be used twice
$$H_{100}\approx \log(201)=\log(3)+\log(67)\approx H_1+H_{33}=\frac{66803685795949}{13127595717600}=5.088...$$
to get an approximation within 2% computing about one third of the sums.
$$\frac{H_{100}}{H_1+H_{33}}=1.019...$$
A slightly worse approximation with about one fifth of the sums is given by:
$$H_{100}=H_{101}-\frac{1}{101}\approx \log(203)-\frac{1}{101}=\log(7)+\log(29)-\frac{1}{101}\approx H_3+H_{14}-\frac{1}{101}=\frac{184711333}{36396360}\approx5.075$$
A: Euler invented some convergence acceleration methods. Probably as an adult, though.
https://en.wikipedia.org/wiki/Series_acceleration
A: Maybe Gauss would have reached
$$H_{100}\approx 1+6\log(2)$$
from
$$\lim_{n \to \infty}\left(\log\left(e^\gamma n+1\right)-H_n\right)=0$$
and
$$e^\gamma\approx\frac{7}{4}$$
so
$$H_n\approx\log\left(\frac{7}{4}n+1\right)$$
For $n=100$, this is
$$H_{100}\approx\log\left(\frac{7}{4}100+1\right)=\log\left(176\right)= 4\log\left(2\right)+\log\left(11\right)$$
but $\log(11)$ is related to $\log(2)$ through the first harmonic number by the same approximation
$$\log(11)=\log(7+4)=\log(7/4+1)+\log(4)\approx H_1-2\log(2)=1+2\log(2)$$
so
$$H_{100}\approx 1+6\log\left(2\right)=5.15888...$$
Now using $\log(2)\approx\frac{2}{3}$ would yield
$$H_{100}\approx 5$$
while $\log(2)\approx\frac{7}{10}$ would lead to
$$H_{100}\approx 5.2$$
