# Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges

I am trying to show if this series converges or diverges and I know it converges since for very large values of n, $$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$ which is convergent from the alternating series test since lim 1/n = 0 and 1/n is a decreasing sequence. I am going back and forth using the limit comparison test and alternating series test and can't seem to come up with anything. So right now i have $$\frac{(-1)^n(n^2+3n-7)}{n^3+1} \lt \frac{(-1)^n(n^2+3n-7)}{n^3}$$ and whenever I try to show the right handed side to be convergent I come back to the same problem as with the left hand side.

• Divide the partial sum of the series in partial fractions, then take all the limits. All parts of the series will converge. Just compare each part with $\sum\frac{(-1)^n}{n}$ – Masacroso Oct 31 '16 at 22:37
• If you have trouble comparing things as $\sum\frac{(-1)^n}{n+k}$ where $k$ is a constant, just change the index, then it will be a subseries of $\sum\frac{(-1)^n}{n}$ or $\sum\frac{(-1)^{n+1}}{n}$. – Masacroso Oct 31 '16 at 22:43
• Comparison tests don't work very well when the sequence is alternating. – Simply Beautiful Art Oct 31 '16 at 23:07
• If the terms shrink fastly enough the (-1)^n is immaterial to convergence. – Jacob Wakem Nov 1 '16 at 1:11

Hint. As $n \to \infty$, we have $$(-1)^n\frac{n^2+3n-7}{n^3+1}=\frac{(-1)^n}n\cdot\frac{1+\frac3n-\frac7{n^2}}{1+\frac1{n^3}}=\frac{(-1)^n}n+O\left(\frac1{n^2} \right)$$ the initial series, being the sum of a conditionally convergent series and an absolute convergent series is then conditionally convergent.

We can write $$\frac{n^2+3n-7}{n^3+1}= \frac{n^2-n +1}{n^3+1} + \frac{4(n +1)}{n^3+1} - \frac{12}{n^3+1}= \\ = \frac{1}{n+1} + \frac{4}{n^2-n +1} - \frac{12}{n^3+1}.$$ Now, $\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n+1}}$ converges conditionally, however, $4\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^2-n +1}}$ and $12\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^3+1}}$ are absolutely convergent.

The alternating or Dirichlet test tells us that if

$$\left|\sum_{n=0}^Na_n\right|<L$$

is bounded and

$$\lim_{n\to\infty}b_n=0$$

where $b_n$ is monotone for all $n>M$, then $\sum_{n=0}^\infty a_nb_n$ converges. In this case, we have $a_n=(-1)^n$ and $b_n=\frac{n^2+3n-7}{n^3+1}$, where $b_n$ is monotone for all $n\ge3$.

• @DonAntonio Thanks for the correction. :) – Simply Beautiful Art Oct 31 '16 at 23:17
• How can you tell it is monotone for n >= 3 – reifi Nov 1 '16 at 2:18
• @reifi You take derivative and find the zeros. – Simply Beautiful Art Nov 1 '16 at 10:56
• @ Simple Art The derivative of b_n is a bit messy isn't it? You just found the zeroes for it? – reifi Nov 1 '16 at 17:04
• @reifi You only need to find an integer value near the root. No need for exactness. – Simply Beautiful Art Nov 1 '16 at 17:09

If you show

$$\left (\frac{x^2+3x-7}{x^3+1} \right)' <0$$

for large $x,$ then for large $n$ the terms of your series are decreasing in absolute value, and you can apply the alternating series test.

• wow that is a surprisingly powerful test. Is the proof simple? (I probably should resolve to prove it myself) – Jacob Wakem Nov 1 '16 at 1:16

I'll take the opportunity for showing some useful techniques for the manipulation of similar series.

Let $\omega=\frac{1+i\sqrt{3}}{2}$ and $\overline{\omega}=\omega^{-1}=\frac{1-i\sqrt{3}}{2}$. By the residue theorem $$\frac{x^2+3x-7}{x^3+1} = \frac{-3}{x+1}+\frac{2+\frac{2i}{\sqrt{3}}}{x-\omega}+\frac{2-\frac{2i}{\sqrt{3}}}{x-\overline{\omega}} \tag{1}$$ where: $$\sum_{n\geq 0}\frac{-3(-1)^n}{n+1} = -3\log(2)\tag{2}$$ so the original series equals $$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} &=& -3\log(2)-4-\sum_{n\geq 0}\frac{(-1)^n 4n}{n^2+n+1}\\&\stackrel{ILP}{=}&-3\log(2)-4-2\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(\frac{s\sqrt{3}}{2}\right)-\sin\left(\frac{s\sqrt{3}}{2}\right)}{\sqrt{3}\cosh\left(\frac{s}{2}\right)}\,ds\\&=&-3\log(2)-4-\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(s\sqrt{3}\right)-\sin\left(s\sqrt{3}\right)}{\cosh\left(s\right)}\tag{3}\end{eqnarray*}$$ where ILP stands for Inverse Laplace Transform. By the Cauchy-Schwarz inequality, the absolute value of the last integral appearing in the RHS of $(3)$ is not larger than $$\frac{8}{\sqrt{3}}\int_{0}^{+\infty}\frac{ds}{\cosh s} = \frac{4 \pi}{\sqrt{3}}.\tag{4}$$ We may also simplify the RHS of $(3)$ a little more and state that: $$\sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} = -\log(8)-4-\frac{2\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}+\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sin(s\sqrt{3})}{\cosh(s)}\,ds.\tag{5}$$