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 shortterms : $\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 
$\displaystyle A_n < \lim_{n\rightarrow \infty } \frac{1}{\sum e^{n}}=1/(e1) <B_n$.


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