How do you determine whether the sequence #a_n=(n+1)^n/n^(n+1)# converges, if so how do you find the limit?

Question

Ezra Adams · Accepted Answer

#lim_(n->oo) a_n = 0#

Note that:

#(1+1/n)^n#

is monotonically increasing with limit #e# as #n->oo#

So:

#lim_(n->oo) a_n = lim_(n->oo) (n+1)^n/(n^(n+1))#

#color(white)(lim_(n->oo) a_n) = lim_(n->oo) ((n+1)/n)^n/n#

#color(white)(lim_(n->oo) a_n) = lim_(n->oo) (1+1/n)^n/n#

#color(white)(lim_(n->oo) a_n) <= lim_(n->oo) e/n#

#color(white)(lim_(n->oo) a_n) = 0#

Axel Alvarez · Answer

undefined To determine whether the sequence $ a_n = \frac{(n+1)^n}{n^{n+1}} $ converges, we can investigate its behavior as $ n $ approaches infinity.

We can rewrite $ a_n $ as:

$$ a_n = \frac{(n+1)^n}{n^{n+1}} = \frac{(1 + \frac{1}{n})^n}{n} $$

The sequence $ a_n $ converges if the limit of $ a_n $ as $ n $ approaches infinity exists.

We can use the fact that $ \lim_{n \to \infty} (1 + \frac{1}{n})^n = e $.

Thus,

$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(1 + \frac{1}{n})^n}{n} = \frac{e}{\infty} = 0 $$

Since the limit of $ a_n $ as $ n $ approaches infinity is 0, the sequence $ a_n $ converges.