Proving Limits of Exponential Series at Infinity

[mathmari]

Hey!

I want to show that $\displaystyle{\lim_{x\rightarrow \infty}\frac{e^x}{x^{\alpha}}=\infty}$ and $\displaystyle{\lim_{x\rightarrow \infty}x^{\alpha}e^{-x}=0}$ using the exponential series (for a fixed $\alpha\in \mathbb{R}$).

I have done the following:

• $$\lim_{x\rightarrow \infty}\frac{e^x}{x^{\alpha}}=\lim_{x\rightarrow \infty}\frac{1}{x^{\alpha}}\cdot e^x = \lim_{x\rightarrow \infty}\frac{1}{x^{\alpha}}\cdot \sum_{n=0}^{\infty}\frac{x^n}{n!} = \lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^n}{x^{\alpha}\cdot n!}= \lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^{n-\alpha}}{ n!}$$
  
  If there are some $n$'s such that $n < \alpha$ then we separate out the finite part of the sum:
  $$\lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^{n-\alpha}}{ n!}=\lim_{x\rightarrow \infty} \left (\sum_{n=0}^{\alpha-1}\frac{x^{n-\alpha}}{ n!}+\frac{x^{\alpha-\alpha}}{ \alpha!}+\sum_{n=\alpha}^{\infty}\frac{x^{n-\alpha}}{ n!}\right )=\lim_{x\rightarrow \infty} \left (\sum_{n=0}^{\alpha-1}\frac{x^{n-\alpha}}{ n!}+\frac{1}{ \alpha!}+\sum_{n=\alpha}^{\infty}\frac{x^{n-\alpha}}{ n!}\right )$$
  
  At the first sum the exponent is negative, and so if $x\rightarrow \infty$ then $x^{n-\alpha}\rightarrow 0$.
  At the second sum the exponent is positive, and so if $x\rightarrow \infty$ then $x^{n-\alpha}\rightarrow \infty$.
  
  Therefore the whole limit goes to $\infty$.
  
  Is everything correct? Could we improve something or make it more formally? (Wondering)
• $$\lim_{x\rightarrow \infty}x^{\alpha}e^{-x}= \lim_{x\rightarrow \infty}x^{\alpha}\cdot \sum_{n=0}^{\infty}\frac{(-x)^n}{n!} = \lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^{\alpha}\cdot (-1)^n\cdot x^n}{ n!}= \lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{(-1)^nx^{n+\alpha}}{ n!}$$
  
  Is everything correct so far? How could we continue? (Wondering)

 

[castor28]

Hi mathmari,

You say that $\alpha\in\mathbb{R}$, but, if $\alpha$ is not an integer, $\alpha!$ is not defined. That is easy to cope with: you can replace $\alpha$ with any integer $k>\alpha$; this will make the expression smaller, and, if it still tends to infinity, you are fine.

If $\alpha\le0$, there is no problem, since the finite part of the sum will disappear.

For the second term, although the general term tends to 0 for any fixed $x$, any particular term tends to infinity, and, as all terms are positive, the sum tends to infinity; this is probably what you mean, but it does no harm to state it explicitly.

For the second part, you could simply observe that the second expression is the inverse of the first one, and use the result of the first part.

 

[mathmari]

You say that $\alpha\in\mathbb{R}$, but, if $\alpha$ is not an integer, $\alpha!$ is not defined. That is easy to cope with: you can replace $\alpha$ with any integer $k>\alpha$; this will make the expression smaller, and, if it still tends to infinity, you are fine.

Do you mean it as follows?

If $\alpha>0$ we let $k$ be the smallest interger such that $k\geq a$, then we have the following:

$$\lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^{n-\alpha}}{ n!}=\lim_{x\rightarrow \infty} \left (\sum_{n=0}^{k}\frac{x^{n-\alpha}}{ n!}+\sum_{n=k+1}^{\infty}\frac{x^{n-\alpha}}{ n!}\right )$$

Or do we split the sum at $k-1$ instead of $k$ ? (Wondering)

For the second term, although the general term tends to 0 for any fixed $x$, any particular term tends to infinity, and, as all terms are positive, the sum tends to infinity; this is probably what you mean, but it does no harm to state it explicitly.

What exactly do you mean by "the general term tends to 0 for any fixed $x$" ? (Wondering)

 

[castor28]

Do you mean it as follows?

If $\alpha>0$ we let $k$ be the smallest interger such that $k\geq a$, then we have the following:

$$\lim_{x\rightarrow \infty} \sum_{n=0}^{\infty}\frac{x^{n-\alpha}}{ n!}=\lim_{x\rightarrow \infty} \left (\sum_{n=0}^{k}\frac{x^{n-\alpha}}{ n!}+\sum_{n=k+1}^{\infty}\frac{x^{n-\alpha}}{ n!}\right )$$

Or do we split the sum at $k-1$ instead of $k$ ?

Hi mathmari,

What I mean is that, if $k>\alpha$, we have:

$$\displaystyle 0 < \frac{e^x}{x^k} < \frac{e^x}{x^\alpha}$$

for $x>1$ (we may assume that $x>1$, since we are interested in what happens for $x\to\infty$).

Therefore, if you can prove that $\dfrac{e^x}{x^k}\to\infty$, the conclusion will follow.

The means that you can completely forget $\alpha$, and prove the proposition for $\dfrac{e^x}{x^k}$; as you can choose $k$ subject ot the condition $k>\alpha$, you may assume that $k$ is an integer, which avoids the problem with the factorials. In practice, this is equivalent to assuming that $\alpha$ is an integer. In particular, you may write $n-k$ instead of $n-\alpha$ for the exponents.

It does not really matter where you split the sum. What is important is that all the terms of the sum are positive, and it is enough to prove that at least one term tends to infinity. This is the case, for example, for the term $\dfrac{x}{(k+1)!}$.

What exactly do you mean by "the general term tends to 0 for any fixed $x$" ?

We can consider the sum as a series in $n$ for $x$ fixed, or as a function of $x$. I just wanted to point out that, for $x$ fixed, $\dfrac{x^{n-k}}{k!}\to0$ as $n\to\infty$ (in fact, we know that the series converges because we know its limit). This is not what we are interested in, however, and I only wanted to avoid any confusion.

 

[castor28]

Hi mathmari,

After thinking a bit more about this, I realize that we may be making it much harder than it needs to be.

We have:

$$\displaystyle
\frac{e^x}{x^\alpha} = \sum_{n=0}^\infty \frac{x^{n-\alpha}}{n!}
$$

The important facts about this series are:

• Like the exponential series, it converges for all $x$.
• If $x>0$ (in particular, if $x\to\infty$), all the terms are positive.

Now, the series contains at least one (actually, infinitely many) term with $n>\alpha$, and this term tends to $+\infty$. As the other terms are positive, they can only make the function grow faster, and we conclude that the function tends to infinity. You don't need to split the series or worry about non-integral or negative values of $\alpha$.

Happy Christmas!
:)

 

[mathmari]

We have:

$$\displaystyle
\frac{e^x}{x^\alpha} = \sum_{n=0}^\infty \frac{x^{n-\alpha}}{n!}
$$

The important facts about this series are:

• Like the exponential series, it converges for all $x$.
• If $x>0$ (in particular, if $x\to\infty$), all the terms are positive.

Now, the series contains at least one (actually, infinitely many) term with $n>\alpha$, and this term tends to $+\infty$. As the other terms are positive, they can only make the function grow faster, and we conclude that the function tends to infinity.

By "As the other terms are positive, they can only make the function grow faster", do you mean as "other terms" the ones with positive exponent, i.e. the ones with $n>\alpha$, or all the other terms including the ones with negative exponent? (Wondering)

About the other limit, can we say something similar, or can we just say the following:

We have:

$$\displaystyle
x^{\alpha}e^{-x}=\frac{x^{\alpha}}{e^x}=\frac{1}{\frac{e^x}{x^\alpha}}
$$

Since $\frac{e^x}{x^\alpha}$ goes to infinity as $x$ goes to infinity, it follows that $\frac{1}{\frac{e^x}{x^\alpha}}$ and so $x^{\alpha}e^{-x}$ goes to $0$.

Happy Christmas!
:)

I wish you also a Merry Christmas! (Smile)

 

[castor28]

Hi mathmari,

By "As the other terms are positive, they can only make the function grow faster", do you mean as "other terms" the ones with positive exponent, i.e. the ones with $n>\alpha$, or all the other terms including the ones with negative exponent?

I mean all the other terms. More precisely, if $k$ is any integer greater than $\alpha$, the sum contains a term $\dfrac{x^{k-\alpha}}{k!}$, which tends to $+\infty$ because the exponent is positive.

As all the terms are positive, we have:

$$\displaystyle
\sum_{n=0}^\infty \frac{x^{n-\alpha}}{n!} > \frac{x^{k-\alpha}}{k!} \to+\infty
$$
​

and this shows that the sum tends to $+\infty$.

About the other limit, can we say something similar, or can we just say the following:

We have:

$$\displaystyle
x^{\alpha}e^{-x}=\frac{x^{\alpha}}{e^x}=\frac{1}{\frac{e^x}{x^\alpha}}
$$

Since $\frac{e^x}{x^\alpha}$ goes to infinity as $x$ goes to infinity, it follows that $\frac{1}{\frac{e^x}{x^\alpha}}$ and so $x^{\alpha}e^{-x}$ goes to $0$.

That is perfectly correct: if $f(x)\to\infty$, then $\dfrac{1}{f(x)}\to0$.

 

[mathmari]

I mean all the other terms. More precisely, if $k$ is any integer greater than $\alpha$, the sum contains a term $\dfrac{x^{k-\alpha}}{k!}$, which tends to $+\infty$ because the exponent is positive.

As all the terms are positive, we have:

$$\displaystyle
\sum_{n=0}^\infty \frac{x^{n-\alpha}}{n!} > \frac{x^{k-\alpha}}{k!} \to+\infty
$$
​

and this shows that the sum tends to $+\infty$.
That is perfectly correct: if $f(x)\to\infty$, then $\dfrac{1}{f(x)}\to0$.

I see! Thank you so much! (Smile)