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. 
 A: 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 $\ln N < 100$.
A: You are right to be worried-the greater than sign is in the wrong direction for what you want to prove.  Your implications also go the wrong way.  You are correct that you want $\ln N \lt 100$, but you should be finding something that implies this, not something that it implies.  Do you have an upper bound for $\ln 2$ or $\ln 10?$  If you know $\ln 10 \lt 2.31$ you are home.
A: Let $H_n$ denotes the $n$-th harmonic number, i.e. $H_n-\sum\limits_{k=1}^n \frac 1k$.
You have
$H_1=1$
$H_3=1+\left(\frac12+\frac13\right)< 1 +2\cdot\frac12 = 2$
$H_7=1+\left(\frac12+\frac13\right)+\left(\frac14+\frac15+\frac16+\frac17\right)< 1+2\cdot\frac12+4\cdot\frac14 = 3$
This leads us to guess that 
$$H_{2^n-1} < n$$
for $n>2$ which can be indeed shown by induction.
Inductive step: If this is true for $n$ then we have 
$$H_{2^{n+1}-1}=H_{2^n-1}+\sum_{2^n-1<k\le 2^{n+1}-1} \frac1k \le H_{2^n-1}+\sum_{2^n-1<k\le 2^{n+1}-1} \frac1{2^n} =
H_{2^n-1} + \frac{2^{n+1}-2^n}{2^n} = H_{2^n-1} + 1 < n+1.$$
Now, as already observed in another answer $10^{30}=1000^{10}<1024^{10}=2^{100}$ so for $N<10^{30}$ you have
$$H_N<H_{2^{100}-1}<100.$$
