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

Question

Ezekiel Alexander · Accepted Answer

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# 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#

Charles Anctil · Answer

See below.

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#.

Scarlett Allison · Answer

undefined To determine if the sequence $ a_n(1 + \frac{1}{\sqrt{n}})^n $ converges and find the limits when they exist, you can use the following steps:

1. First, let's rewrite the sequence as $ a_n \left(1 + \frac{1}{\sqrt{n}}\right)^n $.
2. Then, consider taking the limit as $ n $ approaches infinity: $ \lim_{n \to \infty} a_n \left(1 + \frac{1}{\sqrt{n}}\right)^n $.
3. Apply the limit rules to simplify the expression and evaluate the limit.

If the limit exists and is finite, then the sequence converges. Otherwise, if the limit does not exist or is infinite, the sequence diverges.

Note: The specific value of $ a_n $ is not provided in the question, so the convergence or divergence of the sequence depends on the behavior of $ a_n $ and the term $ \left(1 + \frac{1}{\sqrt{n}}\right)^n $ as $ n $ approaches infinity.