I have this math problem: "determine whether the series converges absolutely, converges conditionally, or diverges."

I can use any method I'd like. This is the series:$$\sum_{n=1}^{\infty}(-1)^n\frac{1}{n\sqrt{n+10}}$$

I though about using a comparison test. But I'm not sure what series I can compare $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n+10}}$ to.

  • 3
    $\begingroup$ Think about $\frac{1}{n^p}$. $\endgroup$ – Daniel Fischer Dec 1 '14 at 16:21
  • $\begingroup$ How does it compare to the Harmonic Series? How about the Alternating Harmonic Series? $\endgroup$ – Ari Dec 1 '14 at 16:21
  • $\begingroup$ @DanielFischer You were right, I deleted my answer. $\endgroup$ – Aaron Maroja Dec 1 '14 at 18:55

Converges absolutely


  • $\begingroup$ is this all right @DanielFischer $\endgroup$ – Learnmore Dec 1 '14 at 17:42
  • 2
    $\begingroup$ Yes, now it's correct. Except the last $<$ is a $=$. $\endgroup$ – Daniel Fischer Dec 1 '14 at 17:44

We can use Leibniz theorem: if a_n is a monotone sequence that converges to 0, then the series $\sum_{n=1}^{\infty}(-1)^n a_n$ converges. Let´s have $a_n=\dfrac{1}{n\sqrt{n+10}}$. So we have to prove that $lim$ an=0 and that it is monotone. If you can show that, you have that it converges

  • $\begingroup$ I haven´t noticed that you also need to show whether it converges absolutely. Then the best way to do it what "learnmore" has written... $\endgroup$ – GorTeX Dec 1 '14 at 21:58

Note that the summand is alternating, and converges monotonically to zero in absolute value. This implies that the series is convergent. The series, $S = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n\sqrt{n+10}}$, is absolutely convergent: $$|S| =|\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n\sqrt{n+10}}| \leq \sum_{n=1}^{\infty} |\frac{1}{n\sqrt{n+10}}| \leq \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}} = \sum_{n=1}^{\infty}n^{-\frac{3}{2}}$$ Since $f(t) := t^{-\frac{3}{2}}$ is positive and decreasing on the set $[1,\infty) \subset \mathbb{R}$, it follows that $\sum_{n=1}^{\infty}f(n)$ converges if and only if $\int_{1}^{\infty}f(t)dt$ converges. Since $\int_{1}^{\infty}f(t)dt = \int_{1}^{\infty}t^{-\frac{3}{2}} = 2$, it follows that the series $\sum_{n=1}^{\infty}n^{-\frac{3}{2}}$ is convergent. Therefore $S$ is absolutely convergent. $\square$

  • $\begingroup$ That answer is not correct. You're not entitled to write $S = \sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n\sqrt{n+10}}$ unless you know the series converges. But that is what you're trying to show! $\endgroup$ – zhw. Mar 23 '18 at 1:23
  • $\begingroup$ Technically I can if we were to consider it as a formal sum, or something like the image of a formal power series through an evaluation homomorphism, but you're right: I should have explained why such a declaration was possible! $\endgroup$ – wpm Mar 23 '18 at 20:08
  • $\begingroup$ What's the absolute value of a formal sum? $\endgroup$ – zhw. Mar 23 '18 at 20:11
  • $\begingroup$ If we wanted to (which in this case, we absolutely don't!!!) we could consider the series as an element of the ring of formal Laurant series over the real numbers with the absolute value induced by the discrete valuation defined by $| \sum_{k=m}^{\infty}a_{k}| = a_{m}$ - but again, that has nothing to do with the problem at hand, and I was sloppy in constructing the solution - I'll edit the solution! $\endgroup$ – wpm Mar 23 '18 at 20:35

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