What is the interval of convergence of #sum_1^oo ( (x+2)^n)/n^n
#?

Question

What is the interval of convergence of  #sum_1^oo ( (x+2)^n)/n^n   
#?

Charles Arocha · Accepted Answer

See below.

The Stirling asymptotic approximation is utilized. #n! approx sqrt(2pin)(n/e)^n# we have for large #n# values #(x+2)^n/n^n approx sqrt(2pi n) ((x+2)/e)^n/(n!) = sqrt(2pi)((x+2)/e)((x+2)/e)^(n-1)/((n-1)! n^(1/2))# or #(x+2)^n/n^n le sqrt(2pi)((x+2)/e)((x+2)/e)^(n-1)/((n-1)!)# and then asymptotically #sum_(n=1)^oo (x+2)^n/n^n le sqrt(2pi)((x+2)/e)e^((x+2)/e)# which is convergent for all #x in RR#

Isabella Ackerman · Answer

undefined The interval of convergence for the series $ \sum_{n=1}^{\infty} \frac{(x+2)^n}{n^n} $ can be found using the Ratio Test. The Ratio Test states that if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L $, then the series converges if $ L < 1 $ and diverges if $ L > 1 $.

In this case, $ a_n = \frac{(x+2)^n}{n^n} $.

Thus, $ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x+2)^{n+1}}{(n+1)^{n+1}} \cdot \frac{n^n}{(x+2)^n} \right| = \left| \frac{(x+2)(n^n)}{(n+1)^{n+1}} \right| $.

Taking the limit as $ n \to \infty $ gives $ \lim_{n \to \infty} \left| \frac{(x+2)(n^n)}{(n+1)^{n+1}} \right| = |x+2| $.

Thus, for the series to converge, we need $ |x+2| < 1 $, which gives us the interval of convergence: $ -3 < x < -1 $.

What is the interval of convergence of #sum_1^oo ( (x+2)^n)/n^n #?