Answer is: option3
\( \frac{1}{3} \)Solution:
To determine the radius of convergence, we evaluate \[ \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| \]
Let \[ a_k=\frac{3^k(x+2)^k}{k+1} \]
Then \[ \left|\frac{a_{k+1}}{a_k}\right| = \left| \frac{3^{k+1}(x+2)^{k+1}}{k+2} \cdot \frac{k+1}{3^k(x+2)^k} \right| \]
\[ = 3|x+2|\cdot\frac{k+1}{k+2} \]
Hence \[ \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| = 3|x+2| \]
The series converges if \[ 3|x+2|<1 \]
So, \[ |x+2|<\frac{1}{3} \]
This gives an interval centered at \(x=-2\) with radius \[ r=\frac{1}{3} \]
