# Prove that the series $\sum_{n=1}^{\infty} \frac{4^n+n^3}{2^n+n!}$ is convergent.

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

• limit comparison? – Lord Shark the Unknown Oct 28 '18 at 10:38
• Thanks for the quick reply. How may I do that? – Lorenzo Fioroni Oct 28 '18 at 10:42
• How about identifying the dominant terms in the numerator and denominator? – Lord Shark the Unknown Oct 28 '18 at 10:44
• They should be 4^n and n! – Lorenzo Fioroni Oct 28 '18 at 10:47
• @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. – gimusi Oct 28 '18 at 11:13

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.

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.

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