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

Nov 1, 2017

For ${a}_{n} > 0$ the sequence

${a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n}$

is convergent if the sequence:

${a}_{n} {e}^{\sqrt{n}}$

is also convergent, and in such case:

${\lim}_{n \to \infty} {a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n} = {\lim}_{n \to \infty} {a}_{n} {e}^{\sqrt{n}}$

#### Explanation:

Given the sequence:

${b}_{n} = {a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n}$

if ${a}_{n} > 0$, consider the sequence obtained taking the logarithm of each term:

${c}_{n} = \ln \left({b}_{n}\right) = \ln \left({a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n}\right)$

Using the properties of logarithms:

${c}_{n} = \ln \left({a}_{n}\right) + n \ln \left(1 + \frac{1}{\sqrt{n}}\right)$

that we can also write as:

${c}_{n} = \ln \left({a}_{n}\right) + \sqrt{n} \ln \frac{1 + \frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}}$

Now use the limit:

${\lim}_{x \to 0} \ln \frac{1 + x}{x} = 1$

which implies that:

${\lim}_{n \to \infty} \ln \frac{1 + \frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}} = 1$

and we have that:

${\lim}_{n \to \infty} {c}_{n} = {\lim}_{n \to \infty} \left(\ln \left({a}_{n}\right) + \sqrt{n}\right) = {\lim}_{n \to \infty} \ln \left({a}_{n}\right) + \ln \left({e}^{\sqrt{n}}\right) = {\lim}_{n \to \infty} \ln \left({a}_{n} {e}^{\sqrt{n}}\right)$

If this last limit exists and is finite, i.e. if:

${\lim}_{n \to \infty} \ln \left({a}_{n} {e}^{\sqrt{n}}\right) = L$

Then, as the exponential and the logarithm are continuous functions:

${\lim}_{n \to \infty} {a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n} = {\lim}_{n \to \infty} {e}^{{c}_{n}} = {e}^{\left({\lim}_{n \to \infty} {c}_{n}\right)} = {e}^{L}$

and:

${e}^{L} = {e}^{\left({\lim}_{n \to \infty} \ln \left({a}_{n} {e}^{\sqrt{n}}\right)\right)} = {\lim}_{n \to \infty} {e}^{\ln} \left({a}_{n} {e}^{\sqrt{n}}\right) = {\lim}_{n \to \infty} {a}_{n} {e}^{\sqrt{n}}$

Thus if the ${a}_{n}$ are positive (or at least positive for $n > N$) we can conclude that the sequence

${a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n}$

is convergent if the sequence:

${a}_{n} {e}^{\sqrt{n}}$

is also convergent, and in such case:

${\lim}_{n \to \infty} {a}_{n} {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n} = {\lim}_{n \to \infty} {a}_{n} {e}^{\sqrt{n}}$

Nov 1, 2017

See below.

#### Explanation:

Assuming that the formulation is

${\lim}_{n \to \infty} {a}_{n} = {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n}$ then

${a}_{n} = {\left(1 + \frac{1}{\sqrt{n}}\right)}^{n} = {\left({\left(1 + \frac{1}{\sqrt{n}}\right)}^{\sqrt{n}}\right)}^{\sqrt{n}}$

so

${\lim}_{n \to \infty} {a}_{n} \approx {\lim}_{n \to \infty} {e}^{\sqrt{n}} = \infty$ so

${a}_{n}$ diverges with $n$.