# Using definition of cauchy to show absolute convergence of a series (specific problem)

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?

• 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. Oct 25, 2022 at 17:55

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.

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