Determine the convergence of the sequence e^(1/n).

[lilypetals]

Homework Statement

Determine whether the sequence converges or diverges. If it converges, find the limit.
a_n = e^1/n

Homework Equations

The limit laws, adapted for sequences.

The Attempt at a Solution

I have the solution; I was just wondering if someone might explain it to me.

I would have initially guessed that the limit of this sequence is 0. Not being sure how to proceed, however, I plugged it into WolframAlpha to see the steps for this particular problem.

lim_n-infinity e^1/n

Using the continuity of e^1/n at n=infinity, write lim_n-infinity e^1/n as e^{lim_n-infinity1/n};

The limit of a quotient is the quotient of the limits:

e^{1/lim_n-infinityn}

The limit of n as n approaches infinity is infinity:

=1.

Now, that makes sense, if I consider that 1/n goes to 0 as n goes to infinity, and thus the power of e would be 0, making the limit equal to one. But I'm not clear on exactly how we got there.

First question: what does "using the continuity of e^1/n at n=infinity mean?

Second question: If lim_n-infinity of n is infinity, why does 1/n become 0?

 

[jhae2.718]

Second question: If lim_n-infinity of n is infinity, why does 1/n become 0?

What does 1/n approach as you increase n?

 

[lilypetals]

0, right?

 

[jhae2.718]

Yes. So when we take \lim_{n \to \infty} \frac{1}{n}, we get \frac{1}{\infty}, which is zero.

 

[pergradus]

Remember that for a series to converge the nth term necessarily must go to zero. (The converse is not true in general - a series whose nth term goes to zero does not necessarily converge.)

Looking at the the nth term of e^1/n as n goes to infinity, you see it is equal to 1, which means this series diverges.

 

[jhae2.718]

Remember that for a series to converge the nth term necessarily must go to zero. (The converse is not true in general - a series whose nth term goes to zero does not necessarily converge.)

Looking at the the nth term of e^1/n as n goes to infinity, you see it is equal to 1, which means this series diverges.

From the original post, we are dealing with a sequence. So, as long as there exists a limit, the sequence converges to that limit. For a series, you are correct.

 

[lilypetals]

Yes. So when we take \lim_{n \to \infty} \frac{1}{n}, we get \frac{1}{\infty}, which is zero.

Okay, I think I get it. Does this translate to other situations as well? Dividing/multiplying by infinity causes the quotient or the product to become 0?

 

[pergradus]

From the original post, we are dealing with a sequence. So, as long as there exists a limit, the sequence converges to that limit. For a series, you are correct.

Doh! Misread the title.

 

[jhae2.718]

Okay, I think I get it. Does this translate to other situations as well? Dividing/multiplying by infinity causes the quotient or the product to become 0?

Multiplying by infinity yields infinity. Dividing by infinity yield zero. (Technically speaking, this is an abuse of notation...)

Then we have indeterminate forms:
\frac{\infty}{\infty}, \frac{0}{0}, 0 \cdot \infty, 1^{\infty}, 0^0, \infty^0, \infty-\infty

 

[lavinia]

Okay, I think I get it. Does this translate to other situations as well? Dividing/multiplying by infinity causes the quotient or the product to become 0?

- No. The right way to look at it is to say that 1/n gets arbitrarily small as n increases. Dividing by infinity actually makes no sense. It is a heuristic.

- If you agree that 1/n converges to 0 then its exponential must converge to 1 because the exponential function is continuous.