Proving that if $N<10^{30}$ then $\sum_{n=1}^{N}\frac{1}{n}<101.$ So, I am asked to prove if $N<10^{30}$ then $$\sum_{n=1}^{N}\frac{1}{n}<101.$$ I am given the information that $2^{10}=1024$ and in the previous part of the question I proved that $$0\leqslant \sum_{n=1}^{N}\frac{1}{n}-\ln N\leqslant 1.$$
So I reasoned as follows, $$\sum_{n=1}^N\frac{1}{n}<1+\ln N$$and so we want $1+\ln N < 101$ which means $$\ln N < 100\Rightarrow N<e^{100}=(2+1/2 + 1/6 + \cdots )^{100}\approx 2^{100}=(2^{10})^{10}=(1024)^{10}\approx (10^3)^{10}=10^{30}.$$
Thus, $N<10^{30}$. 

My main concern is over the approximation sign, which should in fact be a strictly greater sign, which then would mean that even though $\ln N<100$ this does not necessarily mean that $N<10^{30}$ which is what I have to show.
So, would this proof be acceptable, or is there any other way to approach the problem? I am mostly looking for hints.
Thanks. 

You have all the right calculations in exactly the wrong order.  Since $N < 10^{30}$, we also know $N < 10^{30} = 1000^{10} < 1024^{10} = 2^{100} < e^{100}$, therefore $\ln N < 100$.
