13
$\begingroup$

I was playing around and I came up with this product, which I believe to be equal to $\mathrm e^2$.

$$ \prod_{k=0}^\infty \left(1 + \frac{1}{k!}\right) \stackrel{?}{=} \mathrm e^2 $$

After calculating $1000$ terms of this product, I got approximately $7.36431$ (compare: $\mathrm e^2 \approx 7.38906 $, so convergence is very slow if existent).

I tried looking at some product definitions of $\rm e$, but none deal with the product I want.

I know that the product converges since $\sum_{k=0}^\infty 1/k!$ and $\sum_{k=0}^\infty 1/k!^2 $ converge as well.

$\endgroup$
  • $\begingroup$ From the numerical value you indicate, estimating the rest of the product on the LHS, it seems the limit of the LHS should be smaller than $\mathrm e^2$. $\endgroup$ – Did Apr 6 '13 at 14:38
  • $\begingroup$ It should be $LHS < RHS$ ? $\endgroup$ – Inceptio Apr 6 '13 at 14:39
  • $\begingroup$ From the first 30 factors, the decimal representation of the exact result starts $7.36430827236725725637277250963105\ldots$ $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 14:46
5
$\begingroup$

Fairly close, but as your computation shows, not close enough. The logarithm of the "rest" is not hard to estimate, and is far too close to $0$ to make up the gap.

Remark: For small positive $x$, by an alternating series argument or otherwise, we have $\log(1+x)\lt x$. So the logarithm of the infinite product $\displaystyle\prod_{k=n}^\infty \left(1+\frac{1}{k!}\right)$ is less than $\displaystyle\sum_{k=n}^\infty \frac{1}{k!}$.

By bounding the tail with a geometric series, we find this is $\lt \dfrac{n+1}{n\cdot n!}$. The exponential of this, for large $n$, is much too close to $1$. A computation of the partial product to $n=8$ is more than sufficient to show that we cannot have equality.

The convergence of this infinite product is fast, in the same ballpark as the convergence of $\displaystyle \sum\dfrac{1}{k!}$.

$\endgroup$
  • $\begingroup$ Indeed, the log of the rest is less than the rest of the series for $e$ itself (because $\ln(1+x)<x$), hence already $<10^{-7}$ after the first ten summands. $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 14:42
  • $\begingroup$ I had suspected that the actual value was less than $e^2$, but I was sort of hoping it was true. Thanks! $\endgroup$ – George V. Williams Apr 6 '13 at 14:47
1
$\begingroup$

Calculating the first 10 million terms of the product with a working precision of 2000 digits gives the result

  7.36430827236725725637277250963105

At the moment I calculate less terms with a higher working precision but I would be surprised if the result would change a lot.

$\endgroup$
  • 2
    $\begingroup$ No need to use so many factors and so much precision. Using thirty factors and only converting rationals to floating point in the last step is enough. In PARI/GP: prod(k=0,30,1+1/k!)+0. The relative error is less than exp(exp(1)-sum(k=0,30,1/k!)) $\endgroup$ – Hagen von Eitzen Apr 6 '13 at 14:50
  • $\begingroup$ yeah but I think it's nice to see that those bounds for the errors holds. $\endgroup$ – Dominic Michaelis Apr 6 '13 at 14:54
  • $\begingroup$ For more digits see oeis.org/A238695 $\endgroup$ – Vaclav Kotesovec Jul 20 '15 at 9:22

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.