The eighth harmonic number happens to be close to $e$.


$$H_8=\sum_{k=1}^8 \frac{1}{k}=\frac{761}{280}\approx2.71(7)$$

This leads to the almost-integer


Some improvement may be obtained as follows.




$$e\approx H_8\left(1+\frac{1}{80^2}\right)\approx 2.7182818(0)$$ http://mathworld.wolfram.com/eApproximations.html

Equivalently $$ \frac{e}{H_8\left(1+\frac{1}{80^2}\right)} \approx 1.00000000751$$

Q: How can this approximation be obtained from a series?

EDIT: After applying the approximation $$H_n\approx \log(2n+1)$$ (https://math.stackexchange.com/a/1602945/134791) to $$e \approx H_8$$

the following is obtained: $$ e - \gamma-\log\left(\frac{17}{2}\right) \approx 0.0010000000612416$$ $$ e \approx \gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3} +6.12416·10^{-11}$$

  • 9
    $\begingroup$ It's very likely just a numerical coincidence and not likely to be related to any known series. By playing around with numbers like $H_n$ (or $\sqrt{2}$, $\pi$ etc.) you are bound to eventually stumble upon identities like this. It's rearly any deep reason behind it. $\endgroup$
    – Winther
    Jan 9, 2016 at 16:52
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    $\begingroup$ Why is $2\pi+e$ so close to $9$ ? $\endgroup$
    – Lucian
    Jan 9, 2016 at 18:25
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    $\begingroup$ @Lucian "Because" we can get the close approximations $e\approx\frac{19}{7}$ and $\pi\approx \frac{22}{7}$ from their continuous fractions, and $2\pi+e\approx 2\frac{22}{7}+\frac{19}{7}=\frac{44+19}{7}=\frac{63}{7}=9$. (although the integer approximations $e\approx\pi\approx 3$ would suffice). There is also a notable integral for $\pi-\frac{22}{7}$. Is there a similar one for $e-\frac{19}{7}$? That would allow writing $2\pi+e-9$ in closed form. If that is not an answer to "why", at least it would answer the question "how close is $2\pi+e$ to $9$?". $\endgroup$ Jan 9, 2016 at 19:50
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    $\begingroup$ @JaumeOliverLafont: The fact that e is so close to $\dfrac{19}7$ is also an odd coincidence. Euler, its discoverer, passed away on September $18,$ that month being, as its name hints, the seventh of the Roman calendar. $\endgroup$
    – Lucian
    Jan 9, 2016 at 20:04
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    $\begingroup$ There are of course exceptions where such results have simple explanations. My favourite in that regard is $\frac{22}{7}-\pi > 0$ which follows from the nice identity $\frac{22}{7}-\pi = \int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$. However such nice and simple results are not to be expected for every single almost identity and this one seems pretty random. If a simple argument exists explaining it I’d love to see it, but I just don’t think it’s likely to exist. $\endgroup$
    – Winther
    Jan 11, 2016 at 9:46

1 Answer 1


Quesly Daniel obtains $$e\approx \frac{19}{7}$$ from $$\int_0^1 x^2(1-x)^2e^{-x}dx = 14-38e^{-1} \approx 0$$ (see https://www.researchgate.net/publication/269707353_Pancake_Functions)

Similarly, $$\int_0^1 x^2(1-x)^2e^{x}dx = 14e-38 \approx 0$$

The approximation may be refined using the expansion $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$$ so $$\frac{1}{14} \int_0^1 x^2(1-x)^2(e^x-1)dx =e-\frac{163}{60}\approx 0$$ gives the truncation of the series to six terms $$e\approx\frac{163}{60}=\sum_{k=0}^{5}\frac{1}{k!}$$ using the largest Heegner number $163$, and

$$\frac{1}{14} \int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-\frac{761}{280}=e-H_8\approx 0$$

gives $$e\approx H_8$$

Similar integrals relate $e$ to its first four convergents $2$,$3$,$\frac{8}{3}$ and $\frac{11}{4}$.

$$\int_0^1 (1-x)e^x dx = e-2$$ $$\int_0^1 x(1-x)e^x dx = 3-e$$ $$\frac{1}{3}\int_0^1 x^2(1-x)e^x dx=e-\frac{8}{3}$$ $$\frac{1}{4}\int_0^1 x(1-x)^2e^x dx=\frac{11}{4}-e$$

These four formulas are particular cases of Lemma 1 by Henry Cohn in A Short Proof of the Simple Continued Fraction Expansion of e.


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