$\sum\limits_{n=1}^{10000000000000000} \frac{1}{n}$ How does wolfram alpha solve $$\sum\limits_{n=1}^{10000000000000000} \frac{1}{n}\approx 37.4186$$so quickly? It solved it in like 3 seconds is there a equation or something 
 A: Yes, in short there is an equation. When you're summing a continuously differentiable function, you have "Euler Summation",
$$ \sum_{y \leq n \leq x} f(n) = \int_y^x f(t)dt + \int_y^x (t - \lfloor t \rfloor)f'(t)dt + f(x)(\lfloor x\rfloor - x) - f(y)(\lfloor y \rfloor - y).$$
Applied to $f(x) = \frac{1}{x}$, you get
$$\sum_{m \leq n} \frac{1}{m} = \log n + \gamma + O(1/n),$$
where $\gamma$ is the Euler Mascheroni constant.
To check,
$$\ln(10000000000000000) + \gamma = 36.8413615 + 0.5772156649 = 37.4185771649.$$
A: One can prove that
$$\sum_{n=1}^N \frac{1}{n} = \ln(N) + \gamma + O(1/N)$$
where $\gamma$ is the Euler-Mascheroni constant, which is defined by this equation and has been tabulated in quite some detail (it is about $0.577216$). The first term comes from noticing that $\sum_{n=1}^N \frac{1}{n}$ is a Riemann sum for calculating $\int_1^N \frac{1}{y} dy = \ln(N)$. The other two terms are corrections.
From this it follows that
$$\sum_{n=1}^{10^{16}} \frac{1}{n} \approx \ln \left ( 10^{16} \right ) + \gamma$$
where the error is something like $10^{-16}/2$, because one can actually prove that the $O(1/N)$ term is less than $1/(2N)$.
A: we can use the prime numbers in geometric series as follow
$$1+1/2+1/3+1/4+1/5+1/6+1/7+...=1+1/2+1/4+1/8+1/16+...1/3+1/9+1/27+...+1/5+1/25+1/125+...$$
$$1+\frac{1/2}{1-1/2}+\frac{1/3}{1-1/3}+\frac{1/5}{1-1/5}+....$$
