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I have the sequence defined as $$a_n = n^2-e^n.$$ Which alternatives do I have to prove that this sequence, that is, $a_n \leq a_{n+1}$ or $a_n \geq a_{n+1}$ for all $n \in \mathbb{N}$?

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    Define the sequence $b_n = A_n - A_{n+1} = (e - 1)e^n - 2n - 1$ for all integers $n \geq 0$ and use induction to show that $b_n > 0$ for all integers $n \geq 0$. – 2017-01-01

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I'm going to present 2 alternatives:

(1) Let $f: \mathbb{R} \to \mathbb{R}$ be the function $f(x) = x^2 - e^x$. $f^{'} (x) = 2x - e^x < 2x - 2^x \leq 0$ for all $x \geq 1$. Now take $a_n : \mathbb{N} \to \mathbb{R}$ to be the sequence defined as $a_n = f(n)$. Then $$a_n \geq a_{n+1}.$$

(2) Let $a_n: \mathbb{N} \to \mathbb{R}$ such that $a_n = n^2-e^n$. Define the difference sequence as $\Delta_n = a_{n+1}-a_n$. Then,

\begin{align} \Delta_n & = a_{n+1} - a_n \\ & = (n+1)^2 - e^{n+1} - (n^2-e^{n}) \\ & = (n+1)^2-n^2 - e^{n+1} + e^n \\ & = n^2 + 2n + 1 -n^2 - e^n(e-1) \\ & = (2n+1) - e^n(e-1) \\ & < (2n+1) - 2^n(2) \\ & \leq (2n+1) -2(n+1) \\ & = -1. \end{align} so for $n \in \mathbb{N}$ we see that $a_n$ is monotonically decreasing.

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    Is there any other alternative then using the derivate to prove this thing? Thanks in advance! – 2017-01-01
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    i'll update my answer with an alternative approach in a couple of minutes. – 2017-01-01
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    added a different approach :) – 2017-01-01
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    In the 5th equality in your second approach shouldn't you have $(2n+1) - e^n(e-1)$? – 2017-01-01