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Find the smallest integer $N$ such that $x_{m+1} \\leq x_m$ for all $m \\geq N$. 

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The smallest integer is [[0]]

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Let $\\{x_n\\}$ be the sequence given by 

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\\[x_n=n^\\var{k} \\var{t}^n\\]

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$x_n=n^k t^n$ where $k$ is a positive integer and $t$ a real number with $0 < t<1$. Find the smallest integer $N$ such that $(m+1)^k t^{m+1} \\leq m^k t^m$ for all $m \\geq N$. 

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The condition $x_{m+1} \\leq x_m$ is $(m+1)^\\var{k} \\var{t}^{m+1} \\leq m^\\var{k} \\var{t}^m$. This can be shown to be equivalent to $m \\geq \\dfrac{\\var{t}^{1/\\var{k}}}{1-\\var{t}^{1/\\var{k}}}=\\var{c} $. We take $N$ to be the smallest integer $\\geq \\var{c} $, so $N=\\var{n}$. Then $m \\geq \\var{c}$ for all $m \\geq \\var{n}$ and therefore $x_{m+1} \\leq x_m$ for all $m \\geq \\var{n}$.

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