Why is the ratio of the number of terms needed to achieve successive integer values in the harmonic series approximately $e$?

Solution 1:

The $n$th harmonic number $$H_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$$ satisfies $$H_n = \log n + \gamma + \frac{1}{2n} - \frac{1}{12 n^2} + O\left(\frac{1}{n^4}\right),$$ where $$\gamma = 0.5772156649\ldots$$ is the Euler-Mascheroni constant (for a little more about this asymptotic approximation, see this subsection of that article). Even for reasonably small $n$, the $\log n$ term dominates the terms in negative powers of $n$. Those terms are hence negligible for our purposes, and we have $$H_n \approx \log n + \gamma.$$ So, rearranging gives that the number $n_N$ of terms of the harmonic series needed to achieve a sum $H_n \geq N$ is $$n_N \approx \exp(N - \gamma).$$

The estimates this expression furnishes for the values $N$ mentioned in the question are (rounding the values in the last column to the nearest $10^{-3}$): \begin{array}{ccc} N & n_N & \exp(N - \gamma)\\ \hline 2 & 4 & 4.148 \\ 3 & 11 & 11.277 \\ 4 & 31 & 30.655 \\ 5 & 83 & 83.327 \\ \vdots & \vdots & \vdots \\ 15 & 1835421 & 1835420.815 \\ 16 & 4989191 & 4989191.048\\ \end{array} As we can see, this estimate produces good values even for very small values of $N$. (Already for $N = 2$, we need $n_N = 4$ terms, and the dominant correction term here only contributes $\frac{1}{2(4)} = \frac{1}{8}$.) It's probably not hard to write down very good bounds on the error of this estimate (and hence justify discarding the remaining terms in the above expansion), but I don't know offhand what they are.

So, even for reasonably small $N$, the ratio of the number $n_{N + 1}$ of terms of the harmonic series needed to achieve a sum $N + 1$ to the number $n_N$ of terms needed for a sum $N$ is $$\frac{n_{N + 1}}{n_N} \approx \frac{\exp((N + 1) - \gamma)}{\exp(N - \gamma)} = \exp(1) = e,$$ as claimed. It should be easy to use the error estimates mentioned above to show that the limit of this ratio as $N \to \infty$ is indeed $e$ as claimed.

Note by the way that the above argument doesn't use that $N$ is an integer, and so we can just as well use our estimate for noninteger $N$. For example, for $N = 2 \pi$ we need $301$ terms, and the formula gives the estimate $\exp(2 \pi - \gamma) = 300.656\ldots .$

Solution 2:

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, $$\frac{x_{n+1}}{x_n}\rightarrow e.$$