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

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 ?

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  • $\begingroup$ It will work, providing $b_0$ exists. The classic example is $b_n=\frac1n$. $\endgroup$ – Henry Nov 19 '14 at 7:32
  • $\begingroup$ by 'exists' you mean this kind of expression : $ any numerator / (x)$ ? $\endgroup$ – Oleg Nov 19 '14 at 7:33
  • $\begingroup$ There's no reason. Some people just like the number $1$ better than $0$. $\endgroup$ – Jair Taylor Nov 19 '14 at 7:34
  • $\begingroup$ 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 $b_n\to0$. $\endgroup$ – xxx Nov 19 '14 at 7:35
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    $\begingroup$ @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. $\endgroup$ – xxx Nov 19 '14 at 7:42
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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)

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

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