How do you determine if #a_n=(1+n)^(1/n)# converge and find the limits when they exist?

Question

Luke Alvarez · Accepted Answer

The series converge to #=1#

Let #y=(1+n)^(1/n)#

Taking logarithms

#lny=ln((1+n)^(1/n))=1/nln(1+n)#

Therefore,

#y=e^(1/nln(1+n))#

So,

#lim_(n->oo)u_n=lim_(n->oo)e^(1/nln(1+n))#

Apply the limit chain rule

#lim_(n->oo)(1/nln(1+n))=lim_(n->oo)(1/(1+n))# ( by L'Hôpital's rule)

#=0#

Therefore,

#lim_(n->oo)e^(1/nln(1+n))=lim_(n->oo)e^0=1#

Emily Ammerman · Answer

undefined To determine if the sequence $a_n = (1+n)^{\frac{1}{n}}$ converges, we can use the limit comparison test with the sequence $b_n = \frac{n}{n}$.

Let's calculate the limit of the ratio $\frac{a_n}{b_n}$:

$$
\lim_{{n \to \infty}} \frac{(1+n)^{\frac{1}{n}}}{n^{\frac{1}{n}}}
$$

Applying L'Hôpital's rule:

$$
\lim_{{n \to \infty}} \frac{\frac{d}{dn}((1+n)^{\frac{1}{n}})}{\frac{d}{dn}(n^{\frac{1}{n}})}
$$

$$
= \lim_{{n \to \infty}} \frac{\frac{1}{n}(1+n)^{\frac{1}{n}}}{\frac{1}{n}n^{\frac{1}{n}-1}}
$$

$$
= \lim_{{n \to \infty}} \frac{(1+n)^{\frac{1}{n}}}{n^{\frac{1}{n}}}
$$

This limit is equal to 1. Therefore, $a_n$ and $b_n$ have the same behavior as $n \rightarrow \infty$. Since $b_n$ diverges (it's just $1$), $a_n$ also diverges.

So, $a_n$ diverges as $n$ goes to infinity.