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Let $S_{N}:=1+\frac{1}{2}+\cdots+\frac{1}{N}$. Find an estimate for $N$ so that $S_{N}>100$.
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By the idea behind the integral test
$\ln (N+1)<S_{N}<1+\ln N$
Thus, to insure that $S_{N}>100$ pick $\ln (N+1)>100$, that is, $N+1>$ $e^{100} \approx 2.7 * 10^{43}$.
On the other hand, if $1+\ln N<100$, then $S_{N}<100$. Here we can pick any $N<e^{99} \approx 10^{43}$.
by Platinum (129,882 points)

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