Alternating Series , why start at n = 1?

Question

$$\sum_{n=1}^\infty(-1)^nb_n$$

Convergent if $b_{n+1} \le b_n$ and if $\lim b_n = 0$

I'm learning taylor series now , and I'm confused with this alternating series test , I've searched around and this test starts with $n=1$.

Question : Why is it like that , won't starting at $n=0$ achieve the same result ?

Answer 1

You can start at any index $n_0$, the behaviour of terms before $n_0$ has no impact on convergence, provided you still have $b_{n+1} \leq b_n$ for $n \geq n_0$ and $b_n \to 0$.

Example where $b_0$ doesn't exist $ \sum_{n=1}^\infty(-1)^n \frac{1}{n}$

(answer taken from comments)

Answer 2

In my opinion, they are both okay. I prefer to write $$\sum_{n=0}^\infty(-1)^nb_n.$$ if $b_0$ is defininable.