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I already know that the sequence converges as when $n$ approaches $\infty$:

$$\lim_{n \to\infty}\frac{\frac{n^{2}}{n^{3}}+\frac{(-1)^{n}}{n^{3}}}{\frac{2n^{3}}{n^{3}}+\frac{1}{n^{3}}} \;\;\implies\;\;\lim_{n \to \infty}\frac{0+0}{2+0}=\frac{0}{2}=0$$ I want to find out if this sequence is absolutely convergent or not. What I am not sure if I am doing it right. Its the same process, where I take the absolute value of the sequence and I get the same answer: $\lim_{n \to\infty}\frac{0}{2}=0$. BUT I see that I can divide only by $n^{2}$ and I'll get: $$\lim_{n \to\infty}\frac{\frac{n^{2}}{n^{2}}+\frac{(-1)^{n}}{n^{2}}}{\frac{2n^{3}}{n^{2}}+\frac{1}{n^{2}}} \;\;\implies\;\;\lim_{n \to \infty}\frac{1+0}{2n+0} \;\;\implies\;\;\lim_{n \to \infty}\frac{1}{2n}$$

Which is awfully similar to $\frac{1}{n}$ that diverges.

Jean-Claude Arbaut
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Sleepy
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2 Answers2

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For $n\ge 2$ we have $n^2+2(-1)^n\ge 0.$ Hence $$2a_n={2n^2+2(-1)^n\over 2n^3 +1}\ge {n^2\over 2n^3+n^3}={1\over 3n}.$$

Ryszard Szwarc
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Let $a_n=\frac{n^{2}+(-1)^{n}}{2n^{3}+1}.$ Then $a_n \ge 0$ for all $n$. Let $b_n =1/n$, then show that

$$\frac{a_n}{b_n} \to 1/2$$

as $n \to \infty$. Hence there is $N$ such that

$$\frac{a_n}{b_n} \ge 1/4$$

for $n>N.$

Can you proceed ?

Fred
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  • May I ask why $\frac{a_{n}}{b_{n}}≥ 1/4$ ? – Sleepy Mar 24 '22 at 15:02
  • @Sleepy Note, that $\frac{a_n}{b_n}\to\frac{1}{2}$, so there is an $N$ such that $|\frac{a_n}{b_n}-\frac{1}{2}| \leq \frac{1}{4}$, i.e. $\frac{a_n}{b_n} \geq \frac{1}{2}-\frac{1}{4} = \frac{1}{4}$ for all $n \geq N$. – psl2Z Apr 01 '22 at 09:43