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I would like to receive some help with the next problem:

The Problem:

I have the problem with the part of the text from the book i am currently studying from. I am confused about some notations. This is the text:

"Let $a_1$, $a_2$, $a_3$, ... be an array of real numbers. Expresion

$$(1)\sum_{n = 1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdot \cdot \cdot + a_n + \cdot \cdot \cdot,$$

is called infinite real series with general member $a_n$, or shorter real series. Sums

$$s_1 = a_1,$$ $$s_2 = a_1 + a_2,$$ $$\cdot \cdot \cdot \cdot \cdot \cdot \cdot$$ $$s_n = a_1 + a_2 + \cdot \cdot \cdot + a_n,$$ $$\cdot \cdot \cdot \cdot \cdot \cdot \cdot$$

are called partial sums of series (1).

Definition 1

If there is a finite limes $\lim_{n \to \infty} s_n = s$ of the array $(s_n)_{n \in \mathbb{N}}$ of the partial sums of series (1), then we can say that that series converge and that its sum is equal $s$. In that case we write $s = \sum_{n = 0}^{\infty} a_n$. This notation is also used when it is $\lim_{n \to \infty} s_n = \pm \infty$. For the series that don't converge (either because $\lim_{n \to \infty} s_n$ is infinite or because it doesn't exist) we say that the series diverge."

My question:

Please, could you help me understand why in the paragraph before the definition, the sum goes from $n = 1$, but in the definition it goes from $n = 0$? Could you tell me what is the reason behind these diferent notations?

This is just the literal begining of the chapter, so i think that it is important to understand this.

Thank you, for your help and your time!

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    $\begingroup$ Just a typo I think. The sum in the definition should start from $n=1$ to be consistent with the rest of the section. $\endgroup$ – gandalf61 Nov 15 '18 at 12:37
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It's a pure typo. Given that all other sums start with $n=1$, you can easily replace

In that case we write $s=\sum_{n=0}^\infty a_n$

with

In that case we write $s=\sum_{n=1}^\infty a_n$


The thing is that the first few elements of a series are usually not what interests us, in the sense that, so long as $a_0$ is defined, the sum $$\sum_{n=1}^\infty a_n$$

converges if and only if the sum

$$\sum_{n=0}^\infty a_n$$ converges, and the two sums only differ by $a_0$, so we can investigate either one, it doesn't really matter. That's why authors sometimes get sloppy. It's not an excuse, just a reason, but there you go.

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  • $\begingroup$ Interestingly, all other sums in this lesson in my book(in examples, theorems and propositions) begin with $n = 0$. Can it be that $n = 1$ was a typo and that i should always consider that given sum starts form $n = 0$, excpet when it's told differently? $\endgroup$ – MathsLearner Nov 15 '18 at 12:46
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    $\begingroup$ @OgnjenMojovic A given sum always starts from the number that is given (unless it's a typo). In a mathematical text, the starting indices are almost always given. But yeah, if every other example starts the sum at $0$, then probably, the definition should do that too. $\endgroup$ – 5xum Nov 15 '18 at 12:48
  • $\begingroup$ It's confusing me, because when they define series (for the first time in the book) they use sum from $n = 1$, and then in all other instances in this lesson, they use sum from $n = 0$. So, it creates a little trust issue (why defining it that way?). At the end, i guess this notation problem isn't going to be a big problem? $\endgroup$ – MathsLearner Nov 15 '18 at 12:52
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    $\begingroup$ @OgnjenMojovic Like I said, practically any property that's of interest when analyzing series is shared among the two sums (one starting at $0$, one at $1$). So the authors got a little sloppy. You can clearly, from the definition of $\sum_{n=1}^\infty$, see what the sum $\sum_{n=0}^\infty$ should be defined as (i.e., the limit of a sequence of partial sums). $\endgroup$ – 5xum Nov 15 '18 at 12:56
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    $\begingroup$ @OgnjenMojovic Exactly. And it's precisely because the two cases are so similar that they messed up. Because people are sloppy with the easy tasks. You are welcome, and remember, if the answer is useful to you and is what you needed, you can always accept it. $\endgroup$ – 5xum Nov 15 '18 at 13:03

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