Convergence of series with integral inside $\sum_{n=1}^\infty (-1)^{n+1} \int_{n}^{n^2}\frac{dx}{x^6+1}$
First time i see integral inside the series. Need to just prove convergence, so i don't think there is a need to integrate and count it all. Is there any fast methods to work with it?
 A: The series converges absolutely. We have
$$\left|(-1)^n\int_n^{n^2} \frac{1}{x^6+1}\, dx\right| = \int_n^{n^2}\frac{1}{x^6+1}\, dx \le \int_n^{\infty}\frac{1}{x^6}\, dx = \frac{1}{5n^5}.$$
Since $\sum 1/n^5$ converges, our series converges absolutely by the comparison test.
A: To show that the series converges, all we need to show is that the integral $\int_n^{n^2} \frac1{1+x^6}\,dx$ decreases monotonically to $0$ for sufficiently large $n$.  
In fact, it is easy to show that the function $F(n)=\int_n^{n^2}\frac{1}{1+x^6}\,dx$ decreases  monotonically to $0$ for $n\ge 2$.  To see this, differentiate $F(n)$ with respect to $n$ to obtain
$$F'(n)=\frac{2n(1+n^6)-(1+n^{12})}{(1+n^6)(1+n^{12})}\tag1$$
Note that $F'(n)<0$ when $n^{12}-2n^7-2n+1>0$.  For $n\ge 2$, $F'(n)<0$.  And we are done!
A: We have
$$ \int_{n}^{n^2}\frac{dx}{x^6+1}=\int_{0}^{n^2}\frac{dx}{x^6+1}-\int_{0}^{n}\frac{dx}{x^6+1} = \int_{0}^{n}\frac{2x}{x^{12}+1}-\frac{1}{x^6+1}\,dx $$
and since the function $\frac{1}{x^6+1}$ is integrable over $\mathbb{R}^+$ the limit of the LHS/RHS as $n\to +\infty$ is zero.
On the other hand the function $\frac{2x}{x^{12}+1}-\frac{1}{x^6+1}$ has a finite number of zeroes in $\mathbb{R}^+$, hence the LHS/RHS is monotonic from some point on and the series $\sum_{n\geq 1}(-1)^n \int_{n}^{n^2}\frac{dx}{x^2+1}\,dx$ is convergent by Leibniz' rule.
