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An infinite series is $\sum_{n=1}^{\infty} a_{n}$

This expression is equivalent to $\lim_{N->\infty}\sum_{n=1}^{N} a_{n}$

Does this mean that an infinite series is nothing but a limit, rather than an actual sum?

Math12345
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3 Answers3

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Yup. If $s_N=\sum_{n=1}^Na_n$, then $\sum_{n\geq 1}a_n=\lim_{N\to +\infty}s_N$, as you pointed. So you can apply general theorems about sequences to series, if you adapt them accordingly to your needs.

Ivo Terek
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  • So does an infinite series have an infinite amount of terms? – Math12345 May 07 '17 at 21:28
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    An infinite series is the limit of a certain sequence. You don't "add infinite terms", but only analyze the behaviour of the sequence of partial sums. Algebraic manipulations happen to work like we were summing infinite terms, so there's no loss in thinking of it like this. – Ivo Terek May 07 '17 at 21:31
  • What does the limit of N --> +$\infty$ (written in your answer) mean, if the number of terms in the sum is not infinite? – Math12345 May 07 '17 at 21:36
  • It means what it means: just a limit. – Ivo Terek May 07 '17 at 21:37
  • So an infinite series is the limit of a sum as the number of terms in the sum increases infinitely, but there is actually not an infinite amount of terms in the sum? – Math12345 May 07 '17 at 21:39
  • That's the gist of it. But as I said, there is no loss in thinking of it as the sum of infinite terms. – Ivo Terek May 07 '17 at 21:41
  • Why is there no loss in thinking of it as the sum of infinite terms? – Math12345 May 07 '17 at 21:50
  • @Math12345 See the second portion of my answer for why there is loss in thinking of it as the sum of infinite terms. – Simply Beautiful Art May 07 '17 at 22:04
  • From what Ivo Terek from analyzing the behavior of sums of our series we can establish the notion of convergence or divergence if the limit exists – Zophikel May 07 '17 at 23:22
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There are two cases :

$1) $the series converges which means that

$$\lim_{N\to+\infty}\sum_{k=k_0}^Na_k \in\mathbb R (=L)$$

and we write $$\sum_{k=k_0}^{+\infty}a_k=L $$

$2) $the series diverges and it has no sense to write $$\sum_{k=k_0}^{+\infty}a_k. $$

hamam_Abdallah
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To answer the question:

Yes, a series is defined as a limit.

$$\sum_{n=1}^\infty a_n:=\lim_{N\to\infty}\sum_{n=1}^Na_n$$

To answer some of your comments:

No, you cannot think of this as an infinite summation, for it does not obey the rules of normal addition. As I recently showed,

$$\begin{align}0&=\color{#4488ee}{\frac11}-\frac11+\color{#4488ee}{\frac12}-\frac12+\color{#44ee88}{\frac13}-\frac13+\color{#44ee88}{\frac14}-\frac14+\color{orange}{\frac15}-\frac15+\color{orange}{\frac16}-\frac16+\dots\\&\stackrel?=\color{#4488ee}{\frac11+\frac12}-\frac11+\color{#44ee88}{\frac13+\frac14}-\frac12+\color{orange}{\frac15+\frac16}-\frac13+\frac17+\frac18-\frac14+\dots\\&\stackrel?=\frac11+\left(\frac12-\frac11\right)+\frac13+\left(\frac14-\frac12\right)+\frac15+\left(\frac16-\frac13\right)+\frac17+\left(\frac18-\frac14\right)+\dots\\&\stackrel?=\frac11-\frac12+\frac13-\frac14+\frac16-\frac14+\dots\\&\stackrel?=\ln(2)\end{align}$$

But clearly $0\ne\ln(2)$.

Notice however, that if you use the definition with the limit, then it is not possible to do all of the above steps... I'll leave it for readers to find which steps are not possible.

As a side note, there are other interesting ways to define an infinite series. These are called regularizations, and one such example:

$$\text{Abel sum:}\sum_{n=1}^\infty a_n:=\lim_{x\to1^-}\left[\lim_{N\to\infty}\sum_{n=1}^Na_nx^n\right]$$

This can be used to yield funky things like

$$\begin{align}\sum_{n=0}^\infty(-1)^n&=\lim_{x\to1^-}\left[\lim_{N\to\infty}\sum_{n=0}^N(-1)^nx^n\right]\\&=\lim_{x\to1^-}\left[\lim_{N\to\infty}\frac{1-(-x)^{N+1}}{1+x}\right]\\&=\lim_{x\to1^-}\left[\frac1{1+x}\right]\\&=\frac12\end{align}$$

noting that $\lim_{N\to\infty}(-x)^{N+1}=0$ when $|x|<1$, which holds for $x$ close to $1$ on the left hand side.

where the equality is only true after regularization, since the original series does not converge in the normal sense.

Simply Beautiful Art
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