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I'd like to determine the nature of the following serie :

$$\sum_{n\ge 2}\prod_{k=2}^n (2-e^{\frac{1}{k}})$$

Let $u_n = \prod_{k=2}^n (2-e^{\frac{1}{k}})$.

So I "have": $$\begin{aligned} \ln(u_n) &= \sum \ln(2-e^{1/k}) \\& \approx \sum \ln(1-1/k + o(1/k))\\ & \approx \sum 1/k- o(1/k))\\ & \approx -\ln(n) = \ln(1/n)\end{aligned}$$ So I guess that $u_n = \Theta (1/n)$ and so $\sum u_n$ diverge. But all those calculations are not correct since $k$ is not always "big" and we can not sum "$o$" arbitrarily.

Sebastien
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2 Answers2

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Basic idea: $\ln(2-e^h) = -h +O(h^2)$ as $h\to 0.$ So $\ln(2-e^{1/k}) = -1/k +O(1/k^2).$ Summing the latter from $k=2,\dots, n$ gives $-\ln n + a_n,$ where $a_n$ is bounded. Thus the $n$th term in the sum is $\ge c/n$ for some $c>0,c$ independent of $n.$

zhw.
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  • Yeah I agree with the idea which is the same than mine I think. Isn't there a problem summing the terms with $O(1/k^2)$ ? – Sebastien Apr 11 '15 at 11:01
  • A series $\sum a_k$ such that $|a_k|\le C/k^2$ for some constant $C$ is absolutely convergent. – zhw. Apr 11 '15 at 16:40
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By partial summation we have $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\left\lfloor t\right\rfloor \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt $$ where $\left\lfloor t\right\rfloor =t-\left\{ t\right\} $ is the floor function and $\left\{ t\right\} $ is the sawthoot function. Then $$=n\log\left(2-e^{1/n}\right)-2\log\left(2-e^{1/2}\right)-\int_{2}^{n}\frac{e^{1/t}}{t\left(2-e^{1/t}\right)}dt+\int_{2}^{n}\left\{ t\right\} \frac{e^{1/t}}{t^{2}\left(2-e^{1/t}\right)}dt $$ and using $\left\{ t\right\} \geq0 $ we get $$\geq n\log\left(2-e^{1/n}\right)-\int_{2}^{n}\frac{e^{1/t}}{t\left(2-e^{1/t}\right)}dt. $$ Now put in the integral $e^{1/t}=u $. We get $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)\geq n\log\left(2-e^{1/n}\right)-\int_{e^{1/n}}^{e^{1/2}}\frac{1}{\left(2-u\right)\log\left(u\right)}du $$ and now we observe that, if $n $ is sufficiently large, $\frac{1}{\left(2-u\right)\log\left(u\right)} $ get a maximum at $e^{1/2} $ for $u\in\left[e^{1/2},e^{1/n}\right] $, so $$\sum_{k=2}^{n}\log\left(2-e^{1/k}\right)\geq n\log\left(2-e^{1/n}\right)-\frac{2}{\left(2-e^{1/2}\right)}\left(e^{1/2}-e^{1/n}\right)=n\log\left(2-e^{1/n}\right)+O\left(1\right) $$ so finally we get $$\sum_{n\geq2}\prod_{k=2}^{n}\left(2-e^{1/k}\right)\geq\sum_{n\geq2}\left(2-e^{1/n}\right)^{n}e^{O\left(1\right)} $$ and the series diverge because $$\lim_{n\rightarrow\infty}\left(2-e^{1/n}\right)^{n}=e^{-1}. $$

Marco Cantarini
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