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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.

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  • $\begingroup$ 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, 2013 at 14:38
  • $\begingroup$ It should be $LHS < RHS$ ? $\endgroup$
    – Inceptio
    Apr 6, 2013 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, 2013 at 14:46

2 Answers 2

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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 $\displaystyle \sum\dfrac{1}{k!}$.

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  • $\begingroup$ 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, 2013 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, 2013 at 14:47
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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.

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    $\begingroup$ 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, 2013 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, 2013 at 14:54
  • $\begingroup$ For more digits see oeis.org/A238695 $\endgroup$
    – Vaclav Kotesovec
    Jul 20, 2015 at 9:22

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