$\sum_{n=1}^\infty (-1)^{n+1} \int_{n}^{n^2}\frac{dx}{x^6+1}$

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?

  • 1
    $\begingroup$ Perhaps a comparison test would help? Something like $\int_n^{n^2}\frac{\text dx}{x^6}\ge \int_n^{n^2}\frac{\text dx}{x^6+1}$... Since the terms alternate signs, you only need to prove that the terms tend to $0$. $\endgroup$ – abiessu May 28 '18 at 18:26
  • $\begingroup$ Since the term in the front is just changing the sign on every other term involving the integral, you are going to somehow find a function that bounds the integral itself - perhaps not solve it - and show that that bounding choice converges. $\endgroup$ – eSurfsnake May 28 '18 at 18:27

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.


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


Note that $F'(n)<0$ when $n^{12}-2n^7-2n+1>0$. For $n\ge 2$, $F'(n)<0$. And we are done!


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.

  • $\begingroup$ This is not very different from Mark's answer, I just wanted to remark that the eventual monotonicity just follows from the fact that $\frac{1}{x^6+1}$ is a rational function. $\endgroup$ – Jack D'Aurizio May 28 '18 at 19:07

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