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

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!

• Limit-comparison test? Apr 10 '18 at 14:50
• In the first limit, x or n approaches infinity ? Apr 10 '18 at 14:51
• Sorry, it's n. I'll change that Apr 10 '18 at 14:51
• Shark, comparison with what? Apr 10 '18 at 14:52
• How does $|a_n|$ compare to $1/n^4$? Apr 10 '18 at 14:52

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 $\sum \frac1{n^4}$.

• Gimusi. Kindly check my answer . Is this correct? Seems too simple.Thanks:) Apr 10 '18 at 16:12

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.

• What do you mean by the limit is convergent? Apr 10 '18 at 16:40
• @PeterSzilas Yes Peter it seems correct. I suggest only to change "for all $n \in \mathbb{Z^+}$" with for all $n>\bar n$ (that is eventually) $n^4|a_n| \lt B$, but that is a minor point.
– user
Apr 10 '18 at 16:44
• Gimusi .Thanks. Correct me: if b_n =n^4||a_n| is convergent, the b_n is bounded for all(!!) n.Why would you want to change for n>n_0? Apr 10 '18 at 17:32
• Pinklemonade. Please specify, do you mean n^4|a_n|, or the limit of the sum? Apr 10 '18 at 17:34
• @PeterSzilas Oh yes of course, I was thinking to a particular value fixed but if it is the bound it’s ok as you stated, sorry for my confusion.
– user
Apr 10 '18 at 18:38

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