Given a limit value, prove that a given series is absolutely convergent

Question

The question is: If $\lim\limits_{n \to \infty} n^4|a_n|=1$, then show that $\sum_{i=1}^\infty (-1)^{n+1}a_n$ absolutely converges.

What I've got so far: Since the given limit is equal to 1, the series $\sum_{i=1}^\infty n^4|a_n|$ diverges.

To show that the series in the problem absolutely converges, I need to show that $\sum_{i=1}^\infty |a_n|$ converges.

I thought of using the comparison test, but it would be inconclusive if I compared $|a_n|$ with $n^4|a_n|$.

Any help would be appreciated!

Answer 1

Note that

$$\lim\limits_{n \to \infty} n^4|a_n|=\lim\limits_{n \to \infty} \frac{|a_n|}{\frac1{n^4}}=1$$

then

$$\sum_{i=1}^\infty |a_n|$$

converges by limit comparison test with $\sum \frac1{n^4}$.

Answer 2

For any $\epsilon>0$ there is $N$ such that for $n>N$ $n^4|a_n|<1+\epsilon$. We can choose $N$ such that also $\sum_{k=n}^{m}\frac{1}{n^4}<\frac{\epsilon}{1+\epsilon}$, for $n,m>N$, since $\sum_{n=1}^{\infty}\frac{1}{n^4}$ converges.

Therefore, for $n,m>N$ $$\sum_{k=n}^{m}|a_n|\leq(1+\epsilon)\sum_{k=n}^{m}\frac{1}{n^4}<\epsilon$$

Answer 3

Correct me if wrong.

Seems a bit too easy.

Since $\lim_{n \rightarrow \infty} n^4|a_n| =1$ , convergent, it is bounded .

Let $B$ , positive,, real be such a bound.

Hence for all $n \in \mathbb{Z^+} :$

$n^4|a_n| \lt B$ , or $|a_n| \lt B/n^4$.

$\sum |a_n| \lt B \sum 1/n^4.$

By comparison test $\sum |a_n|$ converges.