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May 18th, 2019, 07:00 AM   #1
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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.
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May 21st, 2019, 05:58 AM   #2
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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.
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June 7th, 2019, 04:45 AM   #3
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$\displaystyle A_n < \lim_{n\rightarrow \infty } \frac{1}{\sum e^{-n}}=1/(e-1) <B_n$.
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