Calculate the following limit where $n \in \mathbb{Z}$ and log is to the base $e$ $$\lim_{x\to\infty} \log \prod_{n=2}^{x} \Bigg(1+\frac{1}{n}\Bigg)^{1/n}$$
Asked
Active
Viewed 119 times
-2
-
Please, write us your thoughts about the problem and what you have tried. – Blex Jul 04 '15 at 08:14
-
Notice, that logarithm is a continuous function. – luka5z Jul 04 '15 at 08:49
-
Yes i know that limit can go inside beacause of continuity of log... – user252347 Jul 04 '15 at 08:52
-
It can "go inside" only if the limit $\lim_{x\to\infty}\prod_{n=2}^{x} \Bigg(1+\frac{1}{n}\Bigg)^{1/n}$ exists (if finite). – luka5z Jul 04 '15 at 08:53
-
also it is enough to calculate this product 2^(1/2).3^(1/6).4^(1/12).5^(1/20)............... – user252347 Jul 04 '15 at 08:53
-
is it supposed to have a known closed form? – user251257 Jul 04 '15 at 12:09
-
I get $\approx$ 0.56459970638442432059266770903770496043604 – PM 2Ring Jul 04 '15 at 12:29
3 Answers
2
Hint:
Expressing $\ln\left(1+\dfrac1n\right)$ by Taylor expansion leads to
$$\sum_{n=2}^\infty\frac1n\left(\frac1{n}-\frac1{2n^2}+\frac1{3n^3}\cdots\right)=\sum_{n=2}^\infty \frac1{n^2}-\frac1{2n^3}+\frac1{3n^4}\cdots=\sum_{k=2}^\infty\frac{(-1)^k\zeta(k)}{k-1}.$$
Not really easier.
-
-
As no closed form of the summation of $1/n^3$ is know, you can't really find a exact answer. – Ahmed S. Attaalla Jul 04 '15 at 09:11
-
@AhmedS.Attaalla: there is indeed no closed form for $\zeta(3)$, but that doesn't mean that there is no closed form for a sum of $\zeta$'s. For instance, $\sum(\zeta(n)-1)=1$ and $\sum(\zeta(n)-1)/n=1-\gamma$. – Jul 04 '15 at 09:14
-
As we know the sum of $1/n^2$ is $pi^2/6$ if we're able to come up with a closed form of this summation than we would be able to come up with a closed form for $1/n^3$. – Ahmed S. Attaalla Jul 04 '15 at 09:16
-
-
@AhmedS.Attaalla No closed form is known for the odd powers; the sums for even powers are related to the Bernoulli Numbers. See https://en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function – Jul 04 '15 at 09:24
1
For the confirm that there is no closed form for series found by Yves Daoust, we can use the identity $$\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}\zeta\left(k\right)}{k}=\gamma x+\log\left(\Gamma\left(x+1\right)\right),\,\,-1<x\leq1. $$ Take the derivate to get $$\sum_{k\geq2}\left(-1\right)^{k}x^{k-1}\zeta\left(k\right)=\gamma+\psi\left(x+1\right) $$ hence, assuming $x\neq0 $ $$\sum_{k\geq2}\left(-1\right)^{k}x^{k-2}\zeta\left(k\right)=\frac{\gamma}{x}+\frac{\psi\left(x+1\right)}{x} $$ and now if we integrate from $0 $ to $1 $ $$\sum_{k\geq2}\frac{\left(-1\right)^{k}\zeta\left(k\right)}{k-1}=\int_{0}^{1}\frac{\gamma+\psi\left(x+1\right)}{x}dx $$ and there is no closed form, but only a numerical values ($1.257746...$). See for example here.
Marco Cantarini
- 33,062
- 2
- 47
- 93
-
Do these series for $\gamma$ and $\psi(z+1)$ help? $$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right) $$ http://math.stackexchange.com/a/1591256/134791 $$\psi(z+1)=\sum_{n=1}^\infty \left(-\frac{1}{n}-\frac{1}{n+z} +\sum_{j=n(n-1)+1}^{n(n+1)}\frac{1}{j} \right), z \neq -1,-2,-3,...$$ http://math.stackexchange.com/a/1607855/134791 – Jaume Oliver Lafont Jan 14 '16 at 17:26
1
Hint: You might want first check if the integral $$ \int_2^x \frac{1}{\xi} \log\left( 1+\frac{1}{\xi} \right) \mathrm d \xi $$ remains finite for $x\to\infty$.
Solution:
I got the limit $-\operatorname{Li}_2(-\frac12) = -\sum_{k=1}^\infty \frac{1}{(-2)^k k^2} \approx 0.45$. Not that satisfying...
user251257
- 9,229