# Divergent sequence (Hanym's question at Yahoo! Answers)

MHB Math Helper

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello Hanym,

I suppose you meant $e^{2n}$, not $e^2n$. One way: we have $a_n=(e^{2n} + 6n) ^{1/2}>n^{1/2}$. Suppose $K>0$ then, $n^{1/2}>K\Leftrightarrow n>K^2$. Choosing $n_0=\lfloor K^2\rfloor+1$, if $n\ge n_0$ then $n^{1/2}>K$ and this means that $\displaystyle\lim_{n\to +\infty}n^{1/2}=+\infty$. As a consequence, $\displaystyle\lim_{n\to +\infty}a_n=+\infty$.

Alternatively, you can use well known properties of elementary functions and the Algebra of divergent sequences: $$\lim_{n\to +\infty}a_n=((+\infty)+(+\infty))^{1/2}=(+\infty)^{1/2}=+\infty$$