Question about Summation in calculus.

Question

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?

Answer 1

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.

Answer 2

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.