Problem 32 on page 630 in Larson Calculus 9e.
In Exercies 29-36, test for convergence or divergence, using each test at least one. Identify which test was used.
$$32. \sum \limits_{n=2}^{\infty} \dfrac{1}{n^3-8} $$
when $n=2$ $a_2 = \frac{1}{2^3-8} = \frac{1}{0}$ Uh, not possible to divide by zero.
I know it converges for $\sum \limits_{n=3}^{\infty} \dfrac{1}{n^3-8} = -\dfrac{1}{3}$
But what is *Undefined*$+-\frac{1}{3} = $ ????
From the comments, it is very likely a typo.
Thus, I will present some problems for more clearity.
$$ \sum \limits_{n=0}^{\infty} \dfrac{1}{n^3-8} = -\dfrac{1}{3}$$ Will not converge because at $n = 2$ there is singularity?
Aka, No series can converge if there exist a singularity?
Answer
In this case, as many pointed out, it seems like a typo.
Regarding:
Thus, I will present some problems for more clearity.
$$ \sum \limits_{n=0}^{\infty} \dfrac{1}{n^3-8} = -\dfrac{1}{3}$$ Will not converge because at $n = 2$ there is singularity?
Aka, No series can converge if there exist a singularity?
Notice that
$$ \sum \limits_{n=0}^{\infty} \dfrac{1}{n^3-8} = -\dfrac{1}{3}$$
is just a (no sense) formula, so it cannot converge or diverge.
The question
Does
$$\sum_{k=0}^\infty a_k$$
converge?
It's a shorthand for
Does the sequence
$$\left(\sum_{k=0}^n a_k\right)_{n\in\Bbb N}$$
converge?
In this case
$$\left(\sum_{k=0}^n \frac1{k^3-8}\right)_{n\in\Bbb N}$$
doesn't even makes sense, you have already noted why, so it can't converge nor diverge.
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