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