# Alternating test

I ran into some alternating test somewhere, and I had difficulty applying it directly (It's not an HW and I'm not a student) : $$\sum_{ n \geq 1 } (-1)^n \frac{n^3+n+1}{\sqrt{2n^7+n^5+1}}$$

The convergence to zero of $\displaystyle \frac{n^3+n+1}{\sqrt{2n^7+n^5+1}}$ is obvious since 3 < 7/2 (power comparison) but I can't figure out how to prove easily the decreasing of the sequence...Derivative of underlying function is ugly and quotient or subtraction of successive terms doesn't seem to help...

Still I know how to conclude because (without giving the details) , I found : $$(-1)^n \frac{n^3+n+1}{\sqrt{2n^7+n^5+1}} = \frac{(-1)^n}{\sqrt{2n}}+O(\frac{1}{n^{5/2}})$$

The first term is convergent by the alternating test since $\displaystyle \frac{1}{\sqrt{2n}}$ decrease to zero and $\displaystyle \sum_{n \geq 1 }\frac{1}{n^{5/2}}$ is convergent by p-test (5/2 > 1).

So the initial series is convergent but only conditionnaly since $\displaystyle \frac{n^3+n+1}{\sqrt{2n^7+n^5+1}} \sim \frac{1}{n^{1/2}}$
(diverges by p-test since 1/2 < 1)

Thanks !

• I don't fully understand the question. You know that $(n^3+n+1)/\sqrt{2n^7+n^5+1}$ converges to zero and you have an alternating series, thus you get convergence of the series. What to you mean with "can't figure out how to prove easily the decreasing of the sequence"? You already stated in the beginning of the sentence that the sequence converges to zero. Seems like I am not smart enough to understand you question / goal. – k1next Jul 7 '14 at 9:37
• @sonystarmap To deduce the convergence of $\sum (-1)^n a_n$ with $a_n > 0$, Leibniz' criterion requires that $a_n$ converges monotonically to $0$. – Daniel Fischer Jul 7 '14 at 9:39
• @DanielFischer So the question / goal is to show monotonicity? Thank you! – k1next Jul 7 '14 at 9:40

Now that I understood the question, I think I can help.

The derivative is actually not that hard to calculate since we only need quotient rule and one chain rule for polynomials. I leave this up to you, but in agreement with WolframAlpha, I got \begin{align} \frac{d}{dx}a_n=\frac{2+6 n^2-5 n^4-3 n^5-14 n^6-9 n^7-2 n^9}{2 (1+n^5+2 n^7)^{3/2}} \end{align} with $a_n$ as you sequence.

The denominator is always positive. In the enumerator, except for $2+6n^2$, all other terms are also negative. But since $2<5n^4$ and $6n^2<14n^6$ for all $n\geq 1$, the enumerator is always negative. This yields monotonicity.

• I found this derivative, but as it seemed complicate with some + and - I thought it wasn't the good approach (so I didn't check much and prefered to do an assymptotic developpment). But this 2<5n^4 and 6n^2 < 14n^6 allows to conclude immediately, thanks ! – Dyoann Jul 7 '14 at 9:55