Answers:
Part 1) [tex]\displaystyle \frac{1}{n^2}[/tex] or 1/(n^2) goes in the box.
Part 2) The series converges
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Work Shown:
[tex]\displaystyle L = \lim_{n\to \infty} \frac{a_n}{b_n}\\\\\\\displaystyle L = \lim_{n\to \infty} \frac{\frac{n+3}{n^3-5n+8}}{\frac{1}{n^2}}\\\\\\\displaystyle L = \lim_{n\to \infty} \frac{n+3}{n^3-5n+8}\times\frac{n^2}{1}\\\\\\\displaystyle L = \lim_{n\to \infty} \frac{n^3+3n^2}{n^3-5n+8}\\\\\\[/tex]
[tex]\displaystyle L = \lim_{n\to \infty} \frac{\frac{n^3}{n^3}+\frac{3n^2}{n^3}}{\frac{n^3}{n^3}-\frac{5n}{n^3}+\frac{8}{n^3}}\\\\\\\displaystyle L = \lim_{n\to \infty} \frac{1+\frac{3}{n}}{1-\frac{5}{n^2}+\frac{8}{n^3}}\\\\\\\displaystyle L = \frac{1+0}{1-0+0}\\\\\\\displaystyle L = 1\\\\\\[/tex]
Because L is finite and positive, i.e. [tex]0 < L < \infty[/tex], this means that the original series given and the series
[tex]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}[/tex]
either converge together or diverge together due to the Limit Comparison Test.
But the simpler series is known to converge (p-series test).
Therefore, the original series converges as well.
The two series likely don't converge to the same value, but they both converge nonetheless.
