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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?

pinklemonade
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2 Answers2

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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

user
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  • so it looks like we were wrong (according to my professor). he said that for very large n, $a_n ~ n^{3a}$, which converges when $a < \frac{-1}{3}$ what are your thoughts? – pinklemonade Apr 18 '18 at 13:53
  • @pinklemonade let consider $a=-1$ then $$\frac{n^{5a} + 3n^a}{n^{2a}}=\frac{n^{-5} + 3n^(-1)}{n^{-2}}=\frac1{n^3}+3n$$ it doesn't seem to have not any idea to converge – user Apr 18 '18 at 14:01
  • exactly!! i am going to talk to my professor after class. thanks – pinklemonade Apr 18 '18 at 14:07
  • So apparently, splitting the one series into 2 is not a good approach as we can't make an appropriate conclusion about the sum of the series. Just because $\frac{1}{n^3}$ converges and $3n$ diverges, we cannot conclude that their sum also diverges. – pinklemonade Apr 18 '18 at 14:26
  • @pinklemonade yes in this case we can split and conclude that the series diverges – user Apr 18 '18 at 14:27
  • how do you conclude that? how do you know for sure that their sum will diverge? – pinklemonade Apr 18 '18 at 14:27
  • we can use comparison limit test or simply note that $\frac1{n^3}+3n>3n\to \infty$ – user Apr 18 '18 at 14:28
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.

user284331
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