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 May 18th, 2019, 07:00 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 Compare the functions Compare $\displaystyle A_n$ with $\displaystyle B_n$ for $\displaystyle n\in \mathbb{N}$ $\displaystyle A_n = \frac{1}{e^{-1^2 }+e^{-2^2 }+e^{-3^2}+...+e^{-n^2 }}$ . $\displaystyle B_n =(1+e^{-1})\cdot (1+e^{-2}) \cdot ... \cdot (1+e^{-n})$. In short-terms : $\displaystyle A_n=(\sum_{\lambda=1}^{n} e^{-\lambda^2 })^{-1}\; \;$ and $\displaystyle \; \; B_n=\prod_{\lambda=1}^{n} (1+e^{-\lambda })$ . Last edited by idontknow; May 18th, 2019 at 07:05 AM. May 21st, 2019, 05:58 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 Using $\displaystyle e^{-n^2 }=(1/e^{n})^{n} \;$I arrived to compare $\displaystyle e^{-n^2 }$ with $\displaystyle e^{1+e^{-n}}$ . $\displaystyle e^{1+e^{-n}} >1>e^{-n^2 }\; \;$ for each n . Which gives $\displaystyle A_n < B_n$. Last edited by idontknow; May 21st, 2019 at 06:00 AM. June 7th, 2019, 04:45 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 634 Thanks: 91 \$\displaystyle A_n < \lim_{n\rightarrow \infty } \frac{1}{\sum e^{-n}}=1/(e-1)

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