For what value(s) of a does $\sum_{n=1}^\infty \frac{n^{5a} + 3n^a}{n^{2a}}$ converge?

Question

If I split the fraction into two, I get $n^{3a} + \frac{3}{n^{a}}$. If $a >0$, then term #1 would be very large, and term #2 would be very small. If $a < 0$, term #1 would be very small, and term #2 would be very large. And if $a = 0$, the limit of the series as $n \to \infty$ would be non-zero, which means the series would diverge. So in my opinion there is no value of $a$ that would make the series converge. Does this reasoning sound correct?

Answer 1

Yes it’s correct since we have that

• for $a=0$ the series diverges

• for $a>0 \implies a_n\sim n^{3a}$ and the series diverges

• for $a<0 \implies a_n\sim 3n^{-a}$ and the series diverges

Answer 2

Yes, it goes through. But we can do it like \begin{align*} \sum\dfrac{n^{4a}+3}{n^{a}}, \end{align*} for $a>0$, $\lim_{n\rightarrow\infty}\dfrac{n^{4a}+3}{n^{a}}\geq\lim_{n\rightarrow\infty}n^{3a}=\infty$. Similar reasoning goes for $a<0$. And $a=0$ is clear.