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can you help me proving that this series is convergent, please?

$$\sum_{n=1}^{\infty} \frac{4^n+n^3}{2^n+n!}$$

I've tried to find a bigger convergent series, but with no luck.

Thanks.

Lorenzo

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    $\begingroup$ limit comparison? $\endgroup$ Oct 28, 2018 at 10:38
  • $\begingroup$ Thanks for the quick reply. How may I do that? $\endgroup$ Oct 28, 2018 at 10:42
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    $\begingroup$ How about identifying the dominant terms in the numerator and denominator? $\endgroup$ Oct 28, 2018 at 10:44
  • $\begingroup$ They should be 4^n and n! $\endgroup$ Oct 28, 2018 at 10:47
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    $\begingroup$ @LorenzoFioroni If you are not forced to use direct comparison test, it is more effective use limit comparison (or also ratio test), which finally reduces to a limit evaluation which is in often simpler than find convenient inequalities. $\endgroup$
    – user
    Oct 28, 2018 at 11:13

3 Answers 3

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HINT

The $n!$ factor in the denominator is stronger than others terms, therefore the series seems prone to converge.

In these cases, as suggested by Lord Shark the Unknown, a good and effective way is choose a simple convergent series (e.g. $\sum \frac1{n^2}$) and proceed by limit comparison test.

In that way the study of the convergence reduces to the calculation of a limit.

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Ratio test: $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \frac{4^{n+1}+(n+1)^3}{4^n+n^3}\cdot \frac{2^n+n!}{2^{n+1}+(n+1)!}=4\cdot 0=0<1.$$ Hence the series converges.

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I managed to resolve the question. I initially did like @LordSharkTheUnknown and @Gimusi advised, taking $a(n)=\frac{2*4^n}{n!}$ and $b(n) = \frac{1}{4^n}$. For the limit comparison test, since

$$ \lim_{n \to \infty} \frac{2*4^n*4^n}{n!} = 0$$ and $b(n) $ converges, $\implies a(n)$ converges. Since my initial function is equal or less than $a(n)$, it converges as well.

Than I' ve noticed that also with the ratio test on my bigger series (the one I've called $a(n)$ ) gives $0$ as result, so it converges, but with less effort :)

Thanks to all of you

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