Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? Does $\sum \frac{n^2 + 1}{n^{3.5} -2}$ converge? I think it does. But I cannot find a series to compare. I tried to compare it with $\sum \frac{1}{n^{1.5}}$, but I do not know how.
$$\sum \frac{n^2 + 1}{n^{3.5} -2}?\sum\frac{n^2}{n^{3.5}-2}?\sum\frac{n^2}{n^{3.5}}$$
 A: Compare it to $\frac{2}{n^{1.5}}$. We know that for sufficiently large $n$:
$$n^{1.5} \le n^{3.5}-4$$
$$n^{3.5}+n^{1.5}\le2(n^{3.5}-2)$$
$$\frac{n^2+1}{n^{3.5}-2}\le\frac{2}{n^{1.5}}$$
A: We can use the limit comparison test with $b_n = 1/n^{1.5}$ The limit of $a_n/b_n = 1 < \infty,$ and $\sum b_n$ is a p series that converges, so $\sum a_n$ converges.
A: You can separate this out into two series, $\sum\frac{n^2}{n^{3.5}-2}+\sum\frac1{n^{3.5}-2}$. Then, you should be able to compare the two series with $\sum\frac1{n^{1.5}}$.
A: Beside the good and simple solutions you already received, setting $x=\frac 1 n$ and using Taylor series around $x=0$, you could show that $$\frac{n^2+1}{n^{7/2}-2}=x^{3/2}+x^{7/2}+2 x^5+2 x^7+4 x^{17/2}+O\left(x^{21/2}\right)$$ Back to $n$ $$\frac{n^2+1}{n^{7/2}-2}=\frac{1}{n^{3/2}}+\frac{1}{n^{7/2}}+\frac{2}{n^5}+\frac{2
   }{n^7}+
   \frac{4}{n^{17/2}}+O\left(\frac{1}{n^{21/2}}\right)$$ and then $$\sum_{n=2}^\infty\frac{n^2+1}{n^{7/2}-2}=-10+\zeta \left(\frac{3}{2}\right)+\zeta \left(\frac{7}{2}\right)+2 \zeta (5)+2 \zeta
   (7)+4 \zeta \left(\frac{17}{2}\right)+\cdots$$ which is $\approx 1.8411$ while the exact summation would lead to $\approx 1.8469$.
Using the ratio test and Taylor again, $$u_n=\frac{n^2+1}{n^{7/2}-2}\implies \frac{u_{n+1}}{u_n}=1-\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)$$ which is inconclusive. But, using Raabe's test and Taylor again, $$n\left(\frac{u_n}{u_{n+1}}-1 \right)=\frac{3}{2}+\frac{3}{8 n}+O\left(\frac{1}{n^{3/2}}\right)$$ the limit is $>1$ then convergence.
