How do you find the interval of convergence #Sigma n3^nx^n# from #n=[0,oo)#?

Question

Scarlett Allison · Accepted Answer

The series

#sum_(n=0)^oo n3^nx^n#

is convergent for #x in (-1/3,1/3)# Given the series: #sum_(n=0)^oo n3^nx^n# We can apply the ratio test to determine the interval of values of #x# for which the series is convergent. We then evaluate: #abs (a_(n+1)/a_n) = abs ( ( (n+1)3^(n+1)x^(n+1))/(n3^nx^n)) = 3((n+1)/n)absx# So we have that: #lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) 3((n+1)/n)absx = 3 absx# The series is then absolutely convergent for #absx < 1/3# and divergent for #abs x > 1/3# In the cases where #abs x = 1/3 # the ration test is indecisive and we have to analyze in detail: (i) #x = 1/3# #sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(1/3)^n = sum_(n=0)^oo n = oo# (iI) #x = -1/3# #sum_(n=0)^oo n3^nx^n = sum_(n=0)^oo n3^n(-1/3)^n = sum_(n=0)^oo (-1)^n n # Now if we can consider the partial sums of even order, we have: #s_(2N) = sum_(n=0)^(2N) (-1)^n n = sum_(n=0)^N 2n-(2n-1) = N# while for the partial sums of odd order: #s_(2N+1) = s_(2N) -(2N+1) = N-2N-1 = -N-1# so the series is irregular. In conclusion the series is convergent for #x in (-1/3,1/3)#

Charles Arocha · Answer

undefined To find the interval of convergence for the series $ \sum_{n=0}^{\infty} n3^n x^n $, we use the ratio test.

1. Apply the ratio test:
   $$
   \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)3^{n+1} x^{n+1}}{n3^n x^n} \right|
   $$

2. Simplify the expression:
   $$
   = \lim_{n \to \infty} \left| \frac{(n+1)3^{n+1}}{n3^n} \right| |x|
   $$
   $$
   = \lim_{n \to \infty} \left| \frac{(n+1)}{n} \cdot 3 \right| |x|
   $$
   $$
   = \lim_{n \to \infty} 3 |x|
   $$

3. Apply the limit:
   $$
   \lim_{n \to \infty} 3 |x| = 3 |x|
   $$

4. The series converges if $ 3|x| < 1 $:
   $$
   3|x| < 1 \Rightarrow |x| < \frac{1}{3}
   $$

5. Determine the interval of convergence:
   The series converges absolutely if $ |x| < \frac{1}{3} $.

Therefore, the interval of convergence for the series $ \sum_{n=0}^{\infty} n3^n x^n $ is $ \left(-\frac{1}{3}, \frac{1}{3}\right) $.