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:






  • $\begingroup$ So, basically your question is the following: If $x_n$ is the number of terms in the partial sum of the harmonic series that exceeds $n\in \mathbb{N}$ for the first time, then is it true that $\lim_{n\to \infty}\dfrac{x_{n+1}}{x_n}=e$? Is that correct? $\endgroup$ Dec 29, 2014 at 8:51
  • $\begingroup$ No, $n$ is the number of terms in the partial sum of the harmonic series that exceeds $x$ for the first time, and $\frac{{x+1}_{n_1}}{x_{n_2}}=e$ approximately. $\endgroup$
    – Bretsky
    Dec 29, 2014 at 8:55
  • $\begingroup$ Just read my reformulation once carefully, I think you will see it is the same. The reason I have used notation $x_n$ is to treat it like a sequence and then have some insight about it. $\endgroup$ Dec 29, 2014 at 8:57
  • $\begingroup$ Oh, I see what you mean, sorry I was just confused since you switched the variables from my original question. $\endgroup$
    – Bretsky
    Dec 29, 2014 at 9:00
  • $\begingroup$ Essentially I am wondering why it equals $e$ and if somebody can show me the math that calculates $e$, since I don't see why this formula should make approximations of $e$. $\endgroup$
    – Bretsky
    Dec 29, 2014 at 9:01

2 Answers 2


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 $\exp(2 \pi - \gamma) = 300.656\ldots .$

  • $\begingroup$ How do you turn $Hn≈logn+γ$ into $nN≈exp(N−γ)$ ? That's the only part I don't understand. I am in first year high school, so could you explain that part in a way that would be more clear for me? $\endgroup$
    – Bretsky
    Dec 29, 2014 at 10:42
  • $\begingroup$ It's more or less renaming and solving: If we want to find the approximate number of terms needed for a sum $\geq N$, we're looking for a number $n$ that in particular satisfies $H_n \approx N$. I call the number of terms needed to make a sum $\geq N$ by the name $n_N$, since the answer uses this quantity for more than one value of $N$. Anyway, we replace $H_n$ with $N$ and $n$ with $n_N$ in our formula, giving $N \approx \log n_N + \gamma$, and solving for $n_N$ in this approximation gives $\log n_N = N - \gamma$ and then $n_N = \exp(N - \gamma)$. Very nice observation, by the way. :) $\endgroup$ Dec 29, 2014 at 19:10

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


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