### Quiz 10.7:

Question:

For what values does the series $$\sum \limits _{n=0} ^\infty \frac{nx^n}{4^n(n^2+1)}$$ converge absolutely and conditionally.

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 $$x= -4$$