What is the value of the following infinite product?
$$\left(1+ \frac{1}{1^1}\right) \left(1+ \frac{1}{2^2}\right) \left(1+ \frac{1}{3^3}\right) \cdots $$
Is the value known?
$\begingroup$
$\endgroup$
5
-
4$\begingroup$ It is likely an irrational number between 70/27 and 71/27. Letting T be the tail of the sum n^-n starting with n=4, one has exp(T)70/27 as a tighter upper bound, I think. $\endgroup$– The Masked AvengerCommented Mar 23, 2015 at 16:30
-
5$\begingroup$ Cross-posted to MSE. $\endgroup$– Najib IdrissiCommented Mar 23, 2015 at 16:52
-
11$\begingroup$ Why do you wish to know? Is there some reason why one should expect an answer in closed form? $\endgroup$– Todd TrimbleCommented Mar 23, 2015 at 17:00
-
6$\begingroup$ Value 2.60361190459951423330221282635 is not known to isc.carma.newcastle.edu.au $\endgroup$– Gerald EdgarCommented Mar 23, 2015 at 17:25
-
7$\begingroup$ Your question could use some context. Do you have a reason you want to know the answer, or is this just a curiosity? If you have a good reason your question would likely be accepted. But as has been mentioned, it's quite likely there is no closed form description of your product, so your question may not have an answer, let alone one that is quick to deduce. Being unmotivated puts it closer to recreational math, which is not off-topic here but is treated less gently. $\endgroup$– Ryan BudneyCommented Mar 23, 2015 at 17:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
I'm not sure what the criterium for a full answer is here, so here a technique for $(1+c_k)$ kind of products, turning the infinite product into an infinite sum:
Via telescoping, for friendly $a_n$ and any $m$, we have
$$\lim_{n\to\infty}a_n=a_m+\sum_{n=m}^\infty\left(\dfrac{a_{n+1}}{a_n}-1\right)\,{a_n}.$$
So define
$$a_n:=\prod_{k=1}^{n-1}\left(1+c_k\right)\hspace{.5cm}\implies\hspace{.5cm}\dfrac{a_{n+1}}{a_n}-1=c_n,$$
and then
$$\prod_{n=1}^\infty\left(1+c_n\right) = \lim_{n\to\infty}a_n = \prod_{k=1}^{m-1}\left(1+c_k\right)+\sum_{n=m}^\infty c_n\prod_{k=1}^{n-1}\left(1+c_k\right).$$
For $c_n=\dfrac{1}{n^n}$, that's
$$\frac{1^1+1}{1^1}\,\frac{2^2+1}{2^2}\frac{3^3+1}{3^3}+\sum_{n=4}^\infty\frac{1}{n^n}\prod_{k=1}^{n-1}\left(1+\frac{1}{k^k}\right)=2.603\dots$$
The first term is the lower bound $\frac{70}{27}=2.592\dots$ that's been pointed out in the comment and the remaining sum $\frac{1}{4^4}\dots+\frac{1}{5^5}\dots$ collects some $\mathcal{O}(10^{-2})$.
Truncation of the product after $k=1$ reveals the infinite product is almost two times Sophomore's dream:
$$\prod_{n=1}^\infty\left(1+n^{-n}\right)\approx 2\sum_{n=1}^\infty \frac{1}{n^n}=2.582\dots$$
-
$\begingroup$ can you give me a proof of your third equation. is it true? @NikolajK $\endgroup$ Commented Mar 24, 2015 at 9:54
-
$\begingroup$ The third line is just the $a_n$ from the second plugged into the first. $\endgroup$ Commented Mar 25, 2015 at 9:35