2
$\begingroup$

When I have a sum in the type of $\sum _{n=0}^{\infty }\:c_nx^n$ and for a given value of $x$ the sum is infinity (or converges to infinity? I am not sure if this is correct) does the Sum converges in general?

For example, the $\sum _{n=0}^{\infty }\:1^n= 1+1+1+\dots=+\infty$ does this sum converges and if yes the terminology is, it converges to infinity?

$\endgroup$
1
  • 4
    $\begingroup$ We usually say the latter sum "diverges" as opposed to converges to infinity. $\endgroup$ Jan 17 at 22:09

2 Answers 2

6
$\begingroup$

The sum diverges to infinity - we wouldn't usually say it converges to infinity, and it would certainly be misleading to say it's convergent without further clarification.

The idea of converging is that the sum approaches a limit with smaller and smaller fluctuations: for instance, in $1+1/2+1/4+1/8+\cdots$, the sum is eventually always within $0.1$ of $2$ (after adding five terms), and then eventually always within $0.000000001$, and so on (for any positive $\epsilon$ you choose). The same is true of convergent series that fluctuate above and below the limit, like $1/2-1/4+1/8-1/16+\cdots=1/3$.

In the case of $1+1+1+\cdots$, the sum is never within $0.1$ of $\infty$: each partial sum ($1,2,3,4,\dots$) is infinitely far away from $\infty$. So it doesn't fit the reason we define limits the way we do, nor that defining property of a convergent series.

$\endgroup$
9
  • $\begingroup$ That was what I thought as well however in one of my book when I have the sum; Σ(x^n)= 1/(1-x) for -1<x<1, it says that the sum converges for -1<x<1. He then tests for x=-1 it says it diverges and then for x=1 it says it converges. It says that for x=1 the 1/(x-1) there is a vertical asymptote. It then concludes that the sum converges for -1<x<1, 1. I do not get it. $\endgroup$
    – Vaggelis
    Jan 17 at 22:23
  • $\begingroup$ The geometric series converges if and only if $-1<x<1$, and not for $x=1$, so I'm not too sure what the textbook's point is. Can you post the full context to see if there is something lost in translation by your description? $\endgroup$
    – A.M.
    Jan 17 at 22:29
  • $\begingroup$ It is in greek so I don't think you will understand. Do you want me to post the question I mentioned above for Σ(x^n), with its answer that is on the book and try to translate what the question asks, to see if I misunderstood the question? $\endgroup$
    – Vaggelis
    Jan 17 at 22:32
  • $\begingroup$ Ah, I see. I may not be able to help in this case, other than to reassert that we wouldn't say in English that the series converges when $x = 1$. I couldn't tell you if the Greek language uses different terminology. $\endgroup$
    – A.M.
    Jan 17 at 22:38
  • 2
    $\begingroup$ I read that answer as "The radius of convergence is $1$. The convergence domain is $-1<x<1$." $\endgroup$
    – aschepler
    Jan 17 at 23:15
3
$\begingroup$

Before discussing convergence itself, it is profitable to remember the definition of series.

A numerical sequence $s_{n} = a_{1} + a_{2} + \ldots a_{n}$ (when $a_{k}\in\mathbb{K}\in\{\mathbb{R},\mathbb{C}\})$ converges to $s\in\mathbb{K}$ iff

\begin{align*} (\forall\varepsilon > 0)(\exists n_{\varepsilon}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{\varepsilon} \Rightarrow |s_{n} - s| < \varepsilon) \end{align*}

In order to conclude if it converges or not, there are several approaches which may help to conclude so.

The most known (as far as I know) are the absolute convergence, comparison test, the Leibniz test, the ratio test, the root test, the Cauchy test, the condensation test and so on.

Now consider the following (formal) power series \begin{align*} \sum_{n=0}^{\infty}a_{n}x^{n} := a_{0} + a_{1}x + a_{2}x^{2} + \ldots \end{align*}

Depending on the values of $x$, it may converge or not.

With the purpose of discovering so, you can apply one of the above criteria to study convergence.

Here we present some examples.

Consider the following power series and its corresponding closed form \begin{align*} \sum_{n=0}^{\infty}x^{n} = 1 + x + x^{2} + x^{3} + \ldots = \frac{1}{1-x} \end{align*}

If we are interested in knowing when it converges, we can apply the root test, for instance.

What about the following series? \begin{align*} \sum_{n=0}^{\infty}\frac{x^{n}}{n!} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \ldots \end{align*}

Based on the ratio test, one concludes it converges for every possible real number.

Hence the convergence of a power series strictly depends on the value $x$ you choose to plugin.

Hopefully this helps !

EDIT

With respect to the geometric series, observe the partial sums are given by \begin{align*} s_{n}(x) = 1 + x + x^{2} + x^{3} + \ldots + x^{n-1} \end{align*}

Hence if we multiply both sides by $x$, one arrives at \begin{align*} xs_{n}(x) = x + x^{2} + x^{3} + x^{4} + \ldots + x^{n} \end{align*}

Finally, subtracting the second equation from the first, one gets that \begin{align*} (1-x)s_{n}(x) = 1 - x^{n} \Rightarrow s_{n}(x) = \frac{1 - x^{n}}{1 - x} \end{align*}

So, in order to study the limit of $s_{n}$, it suffices to study the convergence of $x^{n}$.

Let us consider that $|x| < 1$. Then it can be proven (by induction) that $|x|^{n} < |x|^{n-1}$.

Moreover, we do also know that $|x|^{n}\geq 0$. Since $|x|^{n}$ is strictly decreasing and bounded, it converges.

Let us denote such limit by $L$. The next procedure makes us conclude that $L = 0$: \begin{align*} L & = \lim_{n\to\infty}|x|^{n+1}\\\\ & = \lim_{n\to\infty}|x|\times |x|^{n}\\\\ & = |x|\times\lim_{n\to\infty}|x|^{n}\\\\ & = |x|L\\\\ & \Rightarrow L(1 - |x|) = 0\\\\ & \Rightarrow L = 0 \end{align*}

Before answering your question, it is profitable to remind that convergence implies the general term goes to zero as $n$ approaches infinity. Since $|x|^{n}\not\to 0$ whenever $|x|\geq 1$, the convergence of the last series happens iff $|x| < 1$, and we are done.

$\endgroup$
4
  • 1
    $\begingroup$ Could you explain why the downvote? $\endgroup$ Jan 17 at 22:29
  • $\begingroup$ I did not downvote and I find it extremely helpful. For the second sum that you mention why use the root test when theory says that it converges for -1<x<1? And does it converges to 1? If so why? Because there is a question in my book about this sum that says it converges to 1. $\endgroup$
    – Vaggelis
    Jan 17 at 22:36
  • 1
    $\begingroup$ @Vaggelis I have edited my answer. Hopefully it helps ! $\endgroup$ Jan 18 at 2:45
  • 1
    $\begingroup$ Yes it did thank you very much! $\endgroup$
    – Vaggelis
    Jan 18 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.