Solution:
\( \lim \limits _{n \rightarrow \infty} \frac{(n+1)|x^{n+1}|4^n(n^2+1)}{n|x^n|4^{n+1}((n+1)^2+1)} = \frac{|x|}{4}\lim \limits _{n \rightarrow \infty} \frac{(n+1)(n^2+1)}{n(n^2+2n+3)} = \frac{|x|}{4} \lim \limits _{n \rightarrow \infty} \frac{(1+1/n)(1+1/n^2)}{(1+2/n+3/n^2)} = \frac{|x|}{4} \)
Therefore, \( \sum \limits _{n=0} ^\infty \frac{nx^n}{4^n(n^2+1)} \) converges absolutely for \( -4 < x < 4 \) by the Ratio Test
\( x = - 4 \)
\( \sum \limits _{n=0} ^\infty \frac{n(-1)^n}{(n^2+1)} \)
An alternating series with \( u_n= \frac{n}{n^2+1} \)
1) \( u_n>0 \)
2) \( u_n \) non-increasing
3) \( \lim \limits _{n \rightarrow \infty} \frac{n}{n^2+1} = \lim \limits _{n \rightarrow \infty} \frac{1}{n+1/n}=0 \)
So,
\( \sum \limits _{n=0} ^\infty \frac{n(-1)^n}{(n^2+1)} \) converges by AST
\( x = 4 \)
\( \sum \limits _{n=0} ^\infty \frac{n}{(n^2+1)} \)
\( \frac{1}{n},\frac{n}{n^2+1} \) are positive
And, \( \lim \limits _{n \rightarrow \infty} \frac{\frac{n}{n^2+1}}{\frac{1}{n}} = \lim \limits _{n \rightarrow \infty} \frac{n^2}{n^2+1} = \lim \limits _{n \rightarrow \infty} \frac{1}{1+1/n^2} = 1 >0 \)
And, \( \sum \limits _{n=1} ^\infty \frac{1}{n} \) is a divergent p-series \( p = 1 \leq 1 \)
So, \( \sum \limits _{n=0} ^\infty \frac{n}{(n^2+1)} \) diverges by LCT
So our power series, \( \sum \limits _{n=0} ^\infty \frac{nx^n}{4^n(n^2+1)} \) converges absolutely for \( -4 < x < 4 \) and conditionally for \( x= -4\)