Why is the ratio of the number of terms needed to achieve successive integer values in the harmonic series approximately $e$? Consider the harmonic series:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5} + \cdots .$$
It takes $1$ term to achieve a partial sum of $1$, since $1$ is the first number.
It takes $4$ terms to achieve a partial sum of $2$: 
$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4} = 2.08\bar{3}$ is the first partial sum of value at least $2$.
To get to $3$ it takes $11$ terms.
To get to $4$ it takes $31$ terms.
To get to $5$ it takes $83$ terms.
If you consider this equation: $x_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ and replace $x$ with any natural number then: ${(x+1)}_{n*e}$ 
I might not be wording this correctly, but I want to know whether and how it can be shown mathematically why $e$ appears to show up in the ratio of the numbers of terms required for partial sums that take on successive integer values.
More examples
The first partial sum in the harmonic series larger than $15$ is 
$$1+\frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{1835421} > 15$$
The first partial sum larger than 16 is
$$1+\frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{4989191} > 16$$
Note that $\frac{4989191}{1835421}$ is an approximation of $e$.
I've calculated the first five of the quotients of the iterations needed to reach partial sums of integers:
$\frac{4}{1}$
$\frac{11}{4}$
$\frac{31}{11}$
$\frac{83}{31}$
$\frac{227}{83}$

Let's $$1+\frac{1}{2}+\cdots+\frac{1}{x_{n}-1}\lt n$$ and $$1+\frac{1}{2}+\cdots+\frac{1}{x_n}\gt n.$$ If for some $N$, $$\frac{1}{x_n+1}+\cdots+\frac{1}{N}\ge 1,$$ then $x_{n+1}\le N$ and if for some $M$, $$\frac{1}{x_n+1}+\cdots+\frac{1}{M}\lt 1-\frac{1}{x_n},$$ then $x_{n+1}\gt M$. Now it is easy to see that if $$
\int_{x_n+1}^{N+1}\frac{1}{x}dx\ge1\Rightarrow N\ge-1+e(x_n+1)=ex_n+e-1,$$then $\frac{1}{x_n+1}+\cdots+\frac{1}{N}\ge 1$, so $$x_{n+1}\le ex_n+e,$$ and if $$\int_{x_n}^{M}\frac{1}{x}dx\lt1-\frac{1}{x_n}\Rightarrow M\lt e^{1-\frac{1}{x_n}}x_n,$$then $\frac{1}{x_n+1}+\cdots+\frac{1}{M}\lt 1-\frac{1}{x_n},$so $$x_{n+1}\ge e^{1-\frac{1}{x_n}}x_n -1.$$ These bounds show that, indeed, $$\frac{x_{n+1}}{x_n}\rightarrow e.$$
