0
$\begingroup$

I have a series $\sum^\infty_{n=2}\frac{(-1)^n}{\sqrt{n(n-1)^2}}$ that I am trying to prove whether it is absolutely convergent or not. I have approached the problem as follows:

Simplify to $\sum^\infty_{n=2}\frac{(1)}{\sqrt{n^3-2n^2+n}}$

Noted that this is less than $\sum \frac{1}{n}$, which diverges. Inconclusive. Noted that this is greater than $\sum \frac{1}{n^\frac{3}{2}}$ which is convergent. Again inconclusive.

Then, I decided to use the definition of cauchy series to prove that this series is cauchy, thus convergent. So, I took the difference of two arbitrary terms: $\sum^\infty_{k}\frac{(1)}{\sqrt{k^3-2k^2+k}} - \sum^\infty_{n}\frac{(1)}{\sqrt{n^3-2n^2+n}}$ where $k>n$. This becomes $\sum^k_{l=k-n}\frac{(1)}{\sqrt{l^3-2l^2+l}}$. I can see that this will go from a number close to one to zero as we increase the number of terms in the series as well as the difference between $k$ and $n$, so this gives me some intuition that we can find any arbitrarily small difference value. However, I am struggling to make this point rigorous. How can I move forward from here?

$\endgroup$
1
  • 1
    $\begingroup$ Your "simplification" is not a simplification at all. Look at $\displaystyle{\sum\frac 1{(n-1)\sqrt n}}$ and think about the limit comparison test. $\endgroup$ Oct 25, 2022 at 17:55

1 Answer 1

1
$\begingroup$

It is much easier to use the comparison test:

$$\sqrt{n(n-1)^2}>\sqrt{(n-1)^3}=(n-1)^{3/2}$$

$$\Rightarrow \frac{1}{\sqrt{n(n-1)^2}}<\frac{1}{(n-1)^{3/2}}$$

Then

$$\sum_{n=2}^\infty \frac{1}{\sqrt{n(n-1)^2}}<\sum_{n=2}^\infty \frac{1}{(n-1)^{3/2}}=\sum_{n=1}^\infty \frac{1}{n^{3/2}}$$

I think you misapplied the comparison in your initial tries.

$\endgroup$
1
  • 1
    $\begingroup$ You're using regular comparison. Limit comparison does not involve inequalities. $\endgroup$ Oct 25, 2022 at 19:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .