# Alternating Series , why start at n = 1?

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

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 ?

• It will work, providing $b_0$ exists. The classic example is $b_n=\frac1n$. – Henry Nov 19 '14 at 7:32
• by 'exists' you mean this kind of expression : $any numerator / (x)$ ? – Oleg Nov 19 '14 at 7:33
• There's no reason. Some people just like the number $1$ better than $0$. – Jair Taylor Nov 19 '14 at 7:34
• 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 $b_n\to0$. – xxx Nov 19 '14 at 7:35
• @Henry Agreed, but if the question is about why we start at $1$ in general, it's not because $b_0$ is never defined (because someteimes it is). Actually, I find more interesting to notice that you can start anywhere, even at $b_{100}$ if necessary, and sometimes it is: if your sequence is decreasing only after a number of "randomly behaving" terms. Of course you don't start on undefined/nonexistent terms. – xxx Nov 19 '14 at 7:42

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