Understanding the Radius of Convergence of e^x and its Series Expansion

[evinda]

Hello! (Wave)

$$e^x= \sum_{n=0}^{\infty} \frac{x^n}{n!} \forall x \in \mathbb{R}$$

i.e. the radius of convergence of $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ is $+\infty$.

Could you explain me how we deduce that the radius of convergence of $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ is $+\infty$?

Do we use the definition, that is the following?

$$R= \frac{1}{\limsup_{n \to +\infty} \sqrt[n]{|a_n|}}$$

If so how can we find the limit $\limsup \sqrt[n]{\frac{1}{n!}}$ ?

 

[I like Serena]

Hey! (Talking)

Informally, $x^n$ increases polynomically, while $n!$ increases exponentially (Stirling's approximation), so the series of their quotient will converge for any $x$. (Nerd)

 

[Nono713]

As is often helpful with factorials (and square roots) you can try taking the logarithm to find the limit:
$$\log \left ( \limsup \sqrt[n]{\frac{1}{n!}} \right ) = \limsup \log \left ( \sqrt[n]{\frac{1}{n!}} \right ) = \limsup \frac{-\log{n!}}{n}$$
Now observe that:
$$\log{n!} = \log{1} + \log{2} + \cdots + \log{n} \leq n \log{n}$$
Can you conclude?

 

[evinda]

Hey! (Talking)

Informally, $x^n$ increases polynomically, while $n!$ increases exponentially (Stirling's approximation), so the series of their quotient will converge for any $x$. (Nerd)

Could you explain me further why we deduce from the above that the series converges for any $x$? (Thinking)

 

[evinda]

As is often helpful with factorials (and square roots) you can try taking the logarithm to find the limit:
$$\log \left ( \limsup \sqrt[n]{\frac{1}{n!}} \right ) = \limsup \log \left ( \sqrt[n]{\frac{1}{n!}} \right ) = \limsup \frac{-\log{n!}}{n}$$
Now observe that:
$$\log{n!} = \log{1} + \log{2} + \cdots + \log{n} \leq n \log{n}$$
Can you conclude?

So is it as follows?

$- \frac{\log{n!}}{n} \geq - \log n \Rightarrow \limsup \left( - \frac{\log{n!}}{n}\right) \geq \limsup(\log n)=-\liminf \log n$

 

[Dethrone]

I think you could also use the ratio test to show that it converges for all $x$.

 

[Nono713]

So is it as follows?

$- \frac{\log{n!}}{n} \geq - \log n \Rightarrow \limsup \left( - \frac{\log{n!}}{n}\right) \geq \limsup(\log n)=-\liminf \log n$

no... we have that the logarithm of the limit is $\leq \lim_{n \to \infty} - \log{n}$, so the limit is $\leq \lim_{n \to \infty} e^{-\log{n}} = \lim_{n \to \infty} 1/n = 0$. hence the radius of convergence is infinite.

EDIT: forgot inequality signs

 

[I like Serena]

Could you explain me further why we deduce from the above that the series converges for any $x$? (Thinking)

As I said, it is informal. (Angel)
An exponential series increases so much faster than a polynomial series, that both the sequence and the series of their quotient converge.

Formally, as Rido said, you can use the ratio test to prove it. (Wasntme)

 

[evinda]

Formally, as Rido said, you can use the ratio test to prove it. (Wasntme)

$\frac{a_{n+1}}{a_n}= \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=\frac{n!}{(n+1)!}=\frac{1}{n+1} \to 0$

What can we deduce from that? (Thinking)

 

[I like Serena]

$\frac{a_{n+1}}{a_n}= \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=\frac{n!}{(n+1)!}=\frac{1}{n+1} \to 0$

What can we deduce from that? (Thinking)

Shouldn't that be:
$$\frac{a_{n+1}}{a_n}= \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}$$
? (Wondering)

From wikipedia:
[box=yellow]The usual form of the test makes use of the limit
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
The ratio test states that:

• if L < 1 then the series converges absolutely;
• if L > 1 then the series does not converge;
• if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

[/box]
Proof is easy enough, since it basically says that a geometric series will dominate the series. (Emo)

 

[evinda]

Shouldn't that be:
$$\frac{a_{n+1}}{a_n}= \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}$$
? (Wondering)

From wikipedia:
[box=yellow]The usual form of the test makes use of the limit
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
The ratio test states that:

• if L < 1 then the series converges absolutely;
• if L > 1 then the series does not converge;
• if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

[/box]
Proof is easy enough, since it basically says that a geometric series will dominate the series. (Emo)

So is it as follows?$$\left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} \right|=\left| \frac{x^{n+1} n!}{x^n (n+1)!}\right|=\left| \frac{x}{n+1} \right| \to 0<1$$

and so the series converges for all $x \in \mathbb{R}$.

Or have I done something wrong? (Thinking)

 

[evinda]

From wikipedia:
[box=yellow]The usual form of the test makes use of the limit
$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
The ratio test states that:

• if L < 1 then the series converges absolutely;
• if L > 1 then the series does not converge;
• if L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.

[/box]

Also, if we would like to calculate the radius of convergence of $p(x)=-\frac{2x}{1-x^2}= \sum_{n=0}^{\infty} (-2) x^{2n+1}$ we would have the following, right?$$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{(-2) x^{2(n+1)+1}}{(-2) x^{2n+1}} \right|=\left| \frac{x^{2n+3}}{x^{2n+1}}\right|=|x^2|$$

So that the series converges absolutely, it has to hold $|x^2|<1 \Rightarrow 0<x^2<1 \Rightarrow -1<x<1 \Rightarrow |x|<1$.

So do we deduce from $|x|<1$ that the radius of convergence is $1$? (Thinking)

 

[Dethrone]

Also, if we would like to calculate the radius of convergence of $p(x)=-\frac{2x}{1-x^2}= \sum_{n=0}^{\infty} (-2) x^{2n+1}$ we would have the following, right?$$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{(-2) x^{2(n+1)+1}}{(-2) x^{2n+1}} \right|=\left| \frac{x^{2n+3}}{x^{2n+1}}\right|=|x^2|$$

So that the series converges absolutely, it has to hold $|x^2|<1 \Rightarrow 0<x^2<1 \Rightarrow -1<x<1 \Rightarrow |x|<1$.

So do we deduce from $|x|<1$ that the radius of convergence is $1$? (Thinking)

Yep, that is correct.

 

[I like Serena]

So is it as follows?$$\left|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} \right|=\left| \frac{x^{n+1} n!}{x^n (n+1)!}\right|=\left| \frac{x}{n+1} \right| \to 0<1$$

and so the series converges for all $x \in \mathbb{R}$.

Or have I done something wrong? (Thinking)

Yep. (Nod)

Also, if we would like to calculate the radius of convergence of $p(x)=-\frac{2x}{1-x^2}= \sum_{n=0}^{\infty} (-2) x^{2n+1}$ we would have the following, right?$$\left| \frac{a_{n+1}}{a_n} \right|=\left| \frac{(-2) x^{2(n+1)+1}}{(-2) x^{2n+1}} \right|=\left| \frac{x^{2n+3}}{x^{2n+1}}\right|=|x^2|$$

So that the series converges absolutely, it has to hold $|x^2|<1 \Rightarrow 0<x^2<1 \Rightarrow -1<x<1 \Rightarrow |x|<1$.

So do we deduce from $|x|<1$ that the radius of convergence is $1$? (Thinking)

Yep. (Nod)

Small correction: if the series converges absolutely, it has to hold $|x^2|<1$ or $|x^2|=1$.
In the first case it is guaranteed, in the second case it is inconclusive. (Nerd)

 

[evinda]

Yep. (Nod)

So in this case is the radius of converge $+\infty$ ? (Thinking)

Small correction: if the series converges absolutely, it has to hold $|x^2|<1$ or $|x^2|=1$.
In the first case it is guaranteed, in the second case it is inconclusive. (Nerd)

If $|x^2|=1$ then:

$$ \sum_{n=0}^{\infty} (-2) x^{2n+1}=\sum_{n=0}^{\infty} (-2) (x^{2})^nx= \sum_{n=0}^{\infty} (-2) x=-2x \sum_{n=0}^{\infty} 1 \to +\infty $$

And so the series does not converge for $|x^2|=1$ and the radius of convergence is $1$, right?So always if we have a power series that is defined for $x \in (-R,R)$ and we find from the ratio test that $\left| \frac{a_{n+1}}{a_n} \right|<1$ we deduce that the radius of convergence is $R$ ? (Thinking)
Or have I understood it wrong?

 

[Nono713]

Note that if $x$ is complex, $\lvert x^2 \rvert = 1$ does not imply $x^2 = 1$.

 

[evinda]

Note that if $x$ is complex, $\lvert x^2 \rvert = 1$ does not imply $x^2 = 1$.

I see... (Smile)

- - - Updated - - -

If we have a power series of the form $\sum_{n=0}^{\infty} a_n x^n$ , $x \in (0,+\infty) $ and $\lim_{n \to +\infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right|=L<1$, would the radius of convergence be $+\infty$? (Thinking)

And if the power series would be defined for $x \in (0,R)$ or $(-5, R), R>5$ would the radius of convergence be $5$? (Thinking)

 

[I like Serena]

If we have a power series of the form $\sum_{n=0}^{\infty} a_n x^n$ , $x \in (0,+\infty) $ and $\lim_{n \to +\infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right|=L<1$, would the radius of convergence be $+\infty$? (Thinking)

And if the power series would be defined for $x \in (0,R)$ or $(-5, R), R>5$ would the radius of convergence be $5$? (Thinking)

Yes and yes. (Smirk)