= Solution
Apply the <integral bounds for a finite harmonic sum> to the omitted terms from thirteen through one hundred:
$$
\ln\frac{101}{13}<\sum_{n=13}^{100}\frac1n<\ln\frac{100}{12}.
$$
Adding the given $H_{12}=86021/27720$ gives $5.15338\ldots<H_{100}<5.22348\ldots$. Both bounds lie between $5.15$ and $5.25$, so every value they permit rounds to the same two <significant figures>:
$$
\boxed{H_{100}=\sum_{n=1}^{100}\frac1n\simeq5.2\quad\text{to two significant figures}.}
$$
The rounding follows from the bounds, without computing all the intervening terms.
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