Question

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

Answer 1

For `a_n > 0` the sequence

`a_n(1+1/sqrtn)^n`

is convergent if the sequence:

`a_n e^sqrtn`

is also convergent, and in such case:

`lim_(n->oo) a_n(1+1/sqrtn)^n = lim_(n->oo) a_n e^sqrtn`

Explanation:

Given the sequence:

`b_n = a_n(1+1/sqrtn)^n`

if `a_n > 0`, consider the sequence obtained taking the logarithm of each term:

`c_n = ln(b_n) = ln(a_n(1+1/sqrtn)^n)`

Using the properties of logarithms:

`c_n =ln(a_n) +nln(1+1/sqrtn)`

that we can also write as:

`c_n =ln(a_n) +sqrtn ln(1+1/sqrtn)/(1/sqrtn)`

Now use the limit:

`lim_(x->0) ln(1+x)/x = 1`

which implies that:

`lim_(n->oo) ln(1+1/sqrtn)/(1/sqrtn) = 1`

and we have that:

`lim_(n->oo) c_n = lim_(n->oo) ( ln(a_n) + sqrtn) = lim_(n->oo) ln(a_n) +ln(e^sqrtn) = lim_(n->oo) ln(a_n e^sqrtn)`

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

`lim_(n->oo) ln(a_n e^sqrtn) = L`

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

`lim_(n->oo) a_n(1+1/sqrtn)^n = lim_(n->oo) e^(c_n) = e^((lim_(n->oo) c_n)) = e^L`

and:

`e^L = e^(( lim_(n->oo) ln(a_n e^sqrtn) )) = lim_(n->oo) e^ ln(a_n e^sqrtn) = lim_(n->oo) a_n e^sqrtn`

Thus if the `a_n` are positive (or at least positive for `n > N`) we can conclude that the sequence

`a_n(1+1/sqrtn)^n`

is convergent if the sequence:

`a_n e^sqrtn`

is also convergent, and in such case:

`lim_(n->oo) a_n(1+1/sqrtn)^n = lim_(n->oo) a_n e^sqrtn`

Answer 2

See below.

Explanation:

Assuming that the formulation is

`lim_(n->oo) a_n = (1+1/sqrtn)^n` then

`a_n = (1+1/sqrtn)^n = ((1+1/sqrtn)^sqrtn)^sqrtn`

so

`lim_(n->oo)a_n approx lim_(n->oo)e^sqrtn = oo` so

`a_n` diverges with `n`.