I saw on Wikipedia: List of representations of e that

$$e=3+\sum_{k=2}^{\infty}\frac{-1}{k!(k-1)k}$$ It was also mentioned that this identity come from consideration on ways to put upper bound of $e$.

Can anyone give me a hint how we can derive this identity? I have tried to look at its Taylor expansion but that approach seems to fail miserably.

Thanks in advance.


You can derive the following telescoping sum for $k\ge2$:

\begin{align*} \frac{1}{k!}+\frac{1}{k!(k-1)k} &=\frac{1}{k!}\left(1+\frac{1}{(k-1)k}\right)\\ &=\frac{1}{k!}\left(1+\frac{1}{k-1}-\frac{1}{k}\right)\\ &=\frac{k-1}{kk!}+\frac{1}{(k-1)k!}\\ &=\frac{k-1}{kk!}+\frac{1}{(k-1)k(k-1)!}\\ &=\frac{k-1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k(k-1)!}\\ &=\frac{1}{k!}-\frac{1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k!}\\ &=\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!} \end{align*}

Hence, if you go back to the original series representation of $e$ as you did:

\begin{align*} e+\sum_{k=2}^{\infty}\frac{1}{k!k(k-1)} &=2+\sum_{k=2}^{\infty}\frac{1}{k!}+\frac{1}{k!k(k-1)}\\ &=2+\sum_{k=2}^{\infty}\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}\\ &=2+1-\lim_{n\rightarrow\infty}\frac{1}{nn!}=3 \end{align*}


As an alternative approach, we have $$ 3-e=\int_{0}^{1}x(1-x)e^x\,dx \tag{1}$$ by IBP and the RHS of $(1)$ is clearly positive, hence $e<3$.
Since $\int_{0}^{1}x(1-x)\frac{x^k}{k!}\,dx = \frac{1}{k!(k+2)(k+3)}$ we also have $$ 3-e = \sum_{k\geq 0}\frac{1}{k!(k+2)(k+3)}\tag{2} $$ by termwise integration of a Taylor series.

  • 2
    $\begingroup$ @OussamaBoussif: it is indeed equivalent by simple rearrangements, but your approach by creative telescoping for tackling the original series is superior, so (+1) to you. $\endgroup$ – Jack D'Aurizio May 26 '17 at 11:24
  • 2
    $\begingroup$ hmm, I think I get it, +1 for the creative use of integration, it's always nice to see different alternatives ^^ $\endgroup$ – Oussama Boussif May 26 '17 at 11:26
  • 2
    $\begingroup$ It might be interesting to point out that the approximations $e\approx\frac{19}{7}$ and $e\approx\frac{193}{71}$ come from similar integrals, with $x(1-x)$ being replaced by $x^2(1-x)^2$ and $x^3(1-x)^3$. $\endgroup$ – Jack D'Aurizio May 26 '17 at 11:29
  • 1
    $\begingroup$ The general form for descending series that follow your pattern, using the Pochhammer symbol, is $$\sum_{k=0}^\infty \frac{1}{k!(k+2n)_{2n}}$$ $\endgroup$ – Jaume Oliver Lafont May 26 '17 at 21:20
  • 1
    $\begingroup$ @AgalnamedDesire math.stackexchange.com/a/1708366/134791 $\endgroup$ – Jaume Oliver Lafont May 28 '17 at 0:05

An equivalent integral

Plugging a relationship similar to the one used by Jack D'Aurizio, namely $$\frac{1}{k!(k-1)k}=\int_0^1 \frac{x^{k-2}(1-x)}{k!} dx$$ into this integral

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

with non-negative integrand in $(0,1)$ that proves $e<3$, yields the series in the question.

$$\sum_{k=2}^\infty \frac{1}{k!(k-1)k}=3-e$$

This relates the series with the inequality $1+x \leq e^x$.

Similar approximations

This integral is similar to the one used to explain why $e$ is close to the eighth harmonic number.

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

with corresponding series $$e=\frac{761}{280}+\frac{1}{7}\sum_{k=2}^\infty \frac{1}{k!(k+3)(k+4)(k+5)}$$

and $$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$

used to explain the observation by Lucian that $2\pi+e$ is close to $9$.

Another series for $3-e$

Yet another series to prove $e<3$ is related to the integer sequence http://oeis.org/A165457.

$$\frac{1}{e}=\frac{1}{3}+\sum_{k=1}^\infty \frac{1}{(2k+1)!(2k+3)}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.