How to prove $\sum_{n=1}^{\infty}(-1)^na_n$ is conditionally convergent?

Question

Let $f_n(x)=x^n+nx-1$, let $a_n$ denote its unique positive root. Then prove $\sum_{n=1}^{\infty}(-1)^na_n$ is conditionally convergent. Below is my solution.

First, $$a_n(a_n^{n-1}+n)=1,$$ because $a_n^{n-1}+n\to \infty$, so $a_n\to0$.

Then $$na_n=1-a_n^n,$$ so $a_n\sim\frac{1}{n}$. So the asymptotic behavior of $a_n$ is similar to that of $\frac{1}{n}$. Then their convergence should also be similar. I think my idea has some validity, is there any way to prove it rigorously? Or give a counterexample? Thanks！

Answer 1

You are on the right track. Write

$$ (-1)^n a_n = \frac{(-1)^n}{n} + (-1)^{n-1}\frac{a_n^n}{n}. $$

1. The sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n} $ converges by Alternating Series Test.

2. The sum $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{a_n^n}{n}$ converges absolutely by Root Test: $$ \lim_{n\to\infty} \sqrt[n]{\left| (-1)^{n-1}\frac{a_n^n}{n} \right|} = \lim_{n\to\infty} \frac{a_n}{n^{1/n}} = \frac{0}{1} = 0 < 1. $$

Since $S = \sum_{n=1}^{\infty} (-1)^n a_n$ is the sum of two convergent series, $S$ itself also converges.

To show that $S$ does not converge absolutely, note the following general observation:

Observation. Suppose $B = \sum_{n=1}^{\infty} b_n$ converges absolutely and $C = \sum_{n=1}^{\infty} c_n$ converges conditionally. Then $A = \sum_{n=1}^{\infty} (b_n + c_n)$ converges conditionally.

Proof. Assume otherwise $A$ converges absolutely. Then by writing $c_n = (b_n + c_n) - b_n$ and applying the triangle inequality,

$$ \sum_{n=1}^{\infty} |c_n| \leq \sum_{n=1}^{\infty} |b_n + c_n| + \sum_{n=1}^{\infty} |b_n| < \infty, $$

contradicting the assumption that $C$ only converges conditionally.