Representation of $e$ as a descending series

Question

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.

score 21 · Accepted Answer · edited May 26 '17 at 19:06

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*}

score 14 · Answer 2 · answered May 26 '17 at 11:21

14

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.

answered May 26 '17 at 11:21

Jack D'Aurizio

353,855

Nice approach, but it doesn't look like the sum given right? – Oussama Boussif May 26 '17 at 11:23
2

@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. – Jack D'Aurizio May 26 '17 at 11:24
2

hmm, I think I get it, +1 for the creative use of integration, it's always nice to see different alternatives ^^ – Oussama Boussif May 26 '17 at 11:26
2

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 $x^3(1-x)^3$. – Jack D'Aurizio May 26 '17 at 11:29
Well, the technique you presented is very powerful then, I don't know how could you come up with such ideas. Your observation is indeed interesting. ^^ – Oussama Boussif May 26 '17 at 11:32
@Jack D'Aurizio What is "IBP"? – May 26 '17 at 15:31
@AgalnamedDesire: Integration By Parts. – Jack D'Aurizio May 26 '17 at 15:58
1

The general form for descending series that follow your pattern, using the Pochhammer symbol, is $$\sum_{k=0}^\infty \frac{1}{k!(k+2n)_{2n}}$$ – Jaume Oliver Lafont May 26 '17 at 21:20
I was wondering what the appropriate generalization was. Do you have a link or reference? – marty cohen May 26 '17 at 23:27
No, just found it now from the answer and comments. The Pochhammer symbol idea comes from the series for $\frac{22}{7}-\pi$ – Jaume Oliver Lafont May 26 '17 at 23:30
For the general form of integrals, https://arxiv.org/pdf/math/0601660.pdf – Jaume Oliver Lafont May 26 '17 at 23:42
@OussamaBoussif whenever I see $$\sum \frac{a_n}n$$ I first think $$\int_0^1\sum a_nx^{n-1}\ dx$$ – Simply Beautiful Art May 27 '17 at 01:54
@Simply Beautiful Art yeah that's right – Oussama Boussif May 27 '17 at 10:42
@Jack D'Aurizio I had not seen the use of the definite integrals $\int_{0}^{1} x^{n}(1 - x)^{n} e^{x} \mathit{dx}$ to approximate $e$. May you provide a reference? – May 27 '17 at 11:09
1

@AgalnamedDesire https://math.stackexchange.com/a/1708366/134791 – Jaume Oliver Lafont May 28 '17 at 00:05

Jaume Oliver Lafont · Answer 3 · 2017-05-28T15:55:12.667

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.

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

Representation of $e$ as a descending series

3 Answers3

Linked