Determine whether the following series converges or diverges. Justify your answer.

Question

I need some real help with this question. I am asked to determine if$$\sum_{n=1}^\infty\frac{n^2}{(3n-1)^4}$$converges or diverges. I have tried this question for many days already and I am so stressed out from it. So what I did was that I used the integral test and then I used partial fraction decomposition to break it down. Then I tried to integrate the outcome but it was very hard to derive the correct answer. Please, could anyone give me some help?

Answer 1

The series converges because$$\lim_{n\to\infty}\frac{\frac{n^2}{(3n-1)^4}}{\frac1{n^2}}=\frac1{3^4}$$and the series $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$ converges.

Answer 2

Note that

$$\frac{n^2}{(3n-1)^4}\sim\frac1{3^4n^2} $$

then use limit comparison test with $\sum \frac1{n^2}$.

Answer 3

You can make the denominator simpler first, by focusing on the leading term, rather than dealing with all its minor features. In this case, just note that $3n-1 \ge 2n$ for $n\ge 1$. Then $$ \sum_{n=1}^{\infty}\frac{n^2}{(3n-1)^4} \le \sum_{n=1}^{\infty}\frac{1}{16n^2}=\frac{\pi^2}{96}. $$ Clearly it doesn't much matter what's going on down there after the leading order... it could be $(3n - \log n)^4$, or $(3n-\sqrt{n})^4$, or whatever. You want to find a simple bound and reduce the complexity as soon as possible.