Harmonic Series

There are several ways to solve this.

You may use: $\displaystyle H_n \approx \ln (n) + \gamma + \cfrac{1}{2n} - \cfrac{1}{12n^2}$ .

This is already a quite accurate estimate for $H_n$ .

Hence, we have: $\displaystyle H_n \approx \ln (n) + \gamma + \cfrac{1}{2n} - \cfrac{1}{12n^2} \geq 7$

By the fact that as n increases, the value of $H_n$ also increases but in a very slow rate, then, the terms $\cfrac{1}{2n} - \cfrac{1}{12n^2}$ will tend to be zero and hence,

$\ln (n) + \gamma \geq 7 \implies n = \left \lceil e^{7 - \gamma} \right \rceil = \boxed{616}$

---

But i just want to ask, how to solve this kind of equations? (i.e. $\displaystyle H_n \approx \ln (n) + \gamma + \cfrac{1}{2n} - \cfrac{1}{12n^2} \geq 7$ ? See below how I solve this kind of equation. )

$\displaystyle H_n \approx \ln (n) + \gamma + \cfrac{1}{2n} - \cfrac{1}{12n^2} \geq 7 \\ \implies \bigg(12n^2 \ln(n) \bigg) + \bigg(12n^2 \gamma \bigg) + 6n - 1 \geq 84n^2 \\ \bigg(12n^2 \ln(n) \bigg) \geq \bigg( (84 - 12\gamma)n^2\bigg) - 6n + 1$

But I was stocked on the last step. What should I do next?