calculus - How can a sum converge if it has a singularity?

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.