MHB Divergent sequence (Hanym's question at Yahoo Answers)

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    Divergent Sequence
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The sequence defined as an = (e^(2n) + 6n)^(1/2) diverges to infinity as n approaches infinity. It is clarified that the expression should be interpreted as e raised to the power of 2n. By demonstrating that an is greater than n^(1/2), it is shown that the limit of n^(1/2) also approaches infinity. Additionally, using properties of divergent sequences confirms that the limit of an is indeed infinity. Thus, the conclusion is that lim (n→∞) an = +∞.
Fernando Revilla
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Here is the question:

an= (e^2n + 6n) ^1/2

Here is a link to the question:

Find the limit of sequence? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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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: $$\lim_{n\to +\infty}a_n=((+\infty)+(+\infty))^{1/2}=(+\infty)^{1/2}=+\infty$$
 
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