How do you find the radius of convergence of the power series #Sigma x^n/(n!)^(1/n)# from #n=[1,oo)#?

Question

Scarlett Allison · Accepted Answer

The series:

#sum_(n=1)^oo x^n/(n!)^(1/n)#

has radius of convergence #R=1# We can apply the ratio test: #abs(a_(n+1)/a_n) = (x^(n+1)/((n+1)!)^(1/n))/(x^n/(n!)^(1/n)) = abs(x^(n+1)/x^n (n!)^(1/n) / ((n+1)!)^(1/n)) = abs(x)/(n+1)^(1/n)# now consider: # lim_(n->oo) (n+1)^(1/n) = lim_(n->oo) (e^(ln(n+1)))^(1/n)= lim_(n->oo) e^(ln(n+1)/n)# As #e^x# is continuous: # lim_(n->oo) (n+1)^(1/n) = e^(lim_(n->oo) (ln(n+1)/n)) = e^0 = 1# So we have: #lim_(n->oo) abs(a_(n+1)/a_n) = abs(x)# which means that the series is absolutely convergent for #abs(x)<1# and divergent for #abs(x) > 1#, that is the radius of convergence is #R=1#

Charles Arocha · Answer

undefined To find the radius of convergence of the power series Σ(x^n/(n!)^(1/n)) from n=1 to infinity, you can use the ratio test. The ratio test states that if lim |a_{n+1}/a_n| = L as n approaches infinity exists, then the series converges if L < 1 and diverges if L > 1.

Applying the ratio test to the given series:

a_n = x^n/(n!)^(1/n)
a_{n+1} = x^(n+1)/((n+1)!)^(1/(n+1))

Taking the absolute value of the ratio of consecutive terms:

|a_{n+1}/a_n| = |(x^(n+1)/((n+1)!)^(1/(n+1))) / (x^n/(n!)^(1/n))|
              = |(x^(n+1) * (n!)^(1/n) / (x^n * ((n+1)!)^(1/(n+1))))|

Simplify the expression:

|a_{n+1}/a_n| = |x * ((n!)^(1/n) / ((n+1)!)^(1/(n+1))))|

As n approaches infinity, we can analyze the limit of this expression:

lim |a_{n+1}/a_n| = |x * lim (((n!)^(1/n)) / ((n+1)!)^(1/(n+1)))|
                  = |x|

Since the limit is independent of n, the ratio test tells us that the series converges for all x values where |x| < 1. Therefore, the radius of convergence is 1.