Vtyjkyuk

Does the series $sum_n=1^infty (fracn^4n^4+2)^n^5-3$ converge?

I started by using the root test:

$$sqrt[n]a_n=left(fracn^4n^4+2right)^fracn^5-3n=left(1+frac2n^4right)^-fracn^5-3n$$

I don't understand the convergence very well so I'm not sure what to do next? Can I make substitution $n^4=u$ and try to make it converge into $e$ or is that not possible?

Yes, it converges. Note that$$lim_ntoinftyleft(1+frac2n^4right)^n^4=e^2.$$Therefore,$$lim_ntoinftyleft(fracn^4n^4+2right)^n^4=e^-2$$and, since $0<e^-2<1$, the series $sum_n=1^infty(e^-2)^n$ converges. Therefore, the series$$sum_n=1^inftyleft(fracn^4n^4+2right)^n^5$$converges and it is easy to deduce from this that your series converges too.

Note that:
$$lim_ntoinfty left(1+frac2n^4right)^-fracn^5-3n=\
lim_ntoinfty left(1+frac2n^4right)^-n^4+frac3n=\
lim_ntoinfty left(1+frac2n^4right)^-n^4cdot lim_ntoinfty left(1+frac2n^4right)^frac3n=\
left(lim_ntoinfty left(1+frac2n^4right)^fracn^42right)^-2cdot 1=e^-2<1.$$
So, based on the root test the original series converges.

As an alternative to root test we have

$$left(fracn^4n^4+2right)^n^5-3=e^(n^5-3)log left(fracn^4n^4+2right)=e^-(n^5-3)log left(1+frac2n^4right)=$$$$=e^-(n^5-3) left(frac2n^4+O(n^-8)right)=e^-2n+O(n^-3))simfrac1e^2nle frac1n^2$$

therfore the given series converges by limit comparison test with $sum frac1e^2n$ which converges by comparison test with $sum frac1n^2$.