TITLE: Why is $e$ close to $H_8$, closer to $H_8\left(1+\frac{1}{80^2}\right)$ and even closer to $\gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3}$?
QUESTION [7 upvotes]: The eighth harmonic number happens to be close to $e$.
$$e\approx2.71(8)$$
$$H_8=\sum_{k=1}^8 \frac{1}{k}=\frac{761}{280}\approx2.71(7)$$
This leads to the almost-integer
$$\frac{e}{H_8}\approx1.0001562$$
Some improvement may be obtained as follows.
$$e=H_8\left(1+\frac{1}{a}\right)$$
$$a\approx6399.69\approx80^2$$
Therefore
$$e\approx H_8\left(1+\frac{1}{80^2}\right)\approx 2.7182818(0)$$
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:
$$ e - \gamma-\log\left(\frac{17}{2}\right) \approx 0.0010000000612416$$
$$ e \approx \gamma+\log\left(\frac{17}{2}\right) +\frac{1}{10^3} +6.12416·10^{-11}$$

REPLY [2 votes]: 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.