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I have the sequence $a_n = \frac{3^n}{1+3^{2n}}$ and I'm trying to find if it's increasing or decreasing. I used the formula $\frac{a_n}{a_{n+1}}$ to get $\frac{a_n}{a_{n+1}}=\frac{1+3^{2n+2}}{3(1+3^{2n})}\geq 1$. Therefore, the sequence is decreasing, but I'm unsure if this is correct.

User123
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jlb0720us
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  • Don't worry; it is correct. So, what did you get? – José Carlos Santos Sep 17 '21 at 18:20
  • Use \ge to get $\ge$ in math mode – jjagmath Sep 17 '21 at 18:24
  • One thing to note is that $\frac {3^n}{1+3^{2n}} = \frac {3^n\cdot \frac 1{3^n}}{(1+3^{2n})\frac 1{3^n}} = \frac 1{\frac 1{3^n} + 3^n}$. In the denominator the $\frac 1{3^n} + ....$ is decreasing be its not a significant value whereas the $... + 3^n$ is increasing by leaps and bounds so $\frac 1{3^n} + 3^n$ is increasing and $\frac 1{\frac 1{3^n} + 3^n }$ is decreasing. – fleablood Sep 17 '21 at 18:33
  • Or $\frac {1 + 3^{2n+2}}{3(1+3^{2n})} \ge 1 \iff 1+3^{2n+2} \ge 3+ 3^{2n+1})$ (assuming that $1+3^{2n}$ is positive, which it is) $\iff 3^{2n+2}-3^{2n+1} \ge 3-1 \iff 3^{2n+1}(3-1) \ge 2 \iff 3^{2n+1}\cdot 2 \ge 2 \iff 3^{2n+1} > 1$. Which it is. – fleablood Sep 17 '21 at 18:37

1 Answers1

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It is correct.

$\frac{1+3^{2n+2}}{3(1+3^{2n})}=$

$\frac {3^{2n+2} + 1}{3^{2n+1} + 3} =$

$\frac {3^{2n+2} + 9 - 8}{3^{2n+1} + 3} =$

$\frac {3^{2n+2} + 9}{3^{2n+1} + 3} - \frac 8{3^{2n+1} + 3} =$

$3 - \frac 8{3^{2n+1} + 3}$

As $n \ge 0$ we know that $2n+1 \ge 1$ and $3^{2n+1} + 3 \ge 6$ so $\frac 8{3^{2n+1}} + 3 \le \frac 86 = \frac 43 < 2$ so

$3- \frac 8{3^{2n+1} + 3} \ge 3-\frac 86 > 3-2 = 1$.

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Although the following may be easier....

As $3(1+3^{2n}) > 0$ we have

$\frac {1 + 3^{2n+2}}{3(1+3^{2n})} \ge 1 \iff$

$1+3^{2n+2} \ge 3(1+3^{2n}) = 3+ 3^{2n+1} \iff$

$3^{2n+2} - 3\ge 3^{2n+1} - 1\iff$

$3(3^{2n+1} - 1) \ge (3^{2n+1} -1)$. And if we assume $3^{2n+1} - 1 > 0$ which we can so long as $n$ is positive then

$3 \ge 1 \implies 3(3^{2n+1} - 1) \ge (3^{2n+1} -1)$ and we are done...

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Regardless, the intuition ought to be that the power $2n$ in the denominator will overpower the the power $n$ in the numerator.

fleablood
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