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I am trying to prove that the series below converges by the comparison test.

\begin{align} \sum_{n=1}^\infty\frac{2n^3+5^n+5\log(n)}{13-n+8^n} \end{align}

How do I show that:

\begin{align} \frac{2n^3+5^n+5\log(n)}{13-n+8^n} \le 16\left(\frac{5}{8}\right)^n \end{align}

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    $\begingroup$ You can just use limit comparison test $\endgroup$ – Jakobian Jun 5 at 20:53
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    $\begingroup$ Work separately with the numerator and denominator. Get an upper bound on the numerator as a multiple of $5^n$ and a lower bound on the denominator as a multiple of $8^n$. $\endgroup$ – Robert Shore Jun 5 at 21:11
  • $\begingroup$ What would I change -n to in order to make it a multiple of 8𝑛? $\endgroup$ – SmithH Jun 6 at 0:52
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Note for n>1: $$ \dfrac{2n^3 + 5^n + 5\log(n)}{13-n+8^n} \leq \dfrac{2\cdot 5^n + 5^n + 5^n}{\frac{8^n}{2}} = 8\bigg( \frac{5}{8} \bigg)^n $$ and note that $\frac{5}{8} < 1$ so the series will converges by geometric series.

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  • $\begingroup$ What do you change the -n to? $\endgroup$ – SmithH Jun 6 at 0:54
  • $\begingroup$ Note that $8^n - n > 8^n/2$ for $n>1$. Then this implies that $\frac{1}{8^n - n} < \frac{1}{\frac{8^n}{2} }$ $\endgroup$ – user209663 Jun 6 at 15:41
  • $\begingroup$ Thank you so much! One more question, what was your intuition to use 8𝑛/2? I think that is what I am struggling with. i.e If I were to look at the question, what made you approach it this way? $\endgroup$ – SmithH Jun 7 at 8:31
  • $\begingroup$ The goal was to some how express everything in term of exponential functions, $8^n$, and $5^n$ . This way, we can group them together! But when we do that, we need to also make sure the inequality is going in the right direction. In this case, I want to make the denominator smaller. As for how or why I picked $8^n/2$, that just come to playing around with it a little bit. Plug in some values and see what is going on. Plug in $n=1$, and $n=2$ and you can see that the inequality hold. Also note that $8^n$ is an exponential function so it's increasing for $n>1$. $\endgroup$ – user209663 Jun 7 at 13:35
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Simple with asymptotic equivalence of functions:

  • $2n^3+5^n+5\log n\sim_\infty 5^n$,
  • $13-n+8^n\sim_\infty 8^n$, so that

$$\frac{2n^3+5^n+5\log n}{13-n+8^n}\sim_\infty\frac{5^n}{8^n}=\Bigl(\frac 58\Bigr)^{\!n},\quad\text{a convergent geometric series.}$$

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