Does this series converge as $x\to \infty$ $$\lim_{x \rightarrow \infty}  \sum_{n=0}^\infty (-1)^{n} \frac{x \lvert x \rvert^{2n}}{2n+1}$$
If the series didn't have  $(-1)^n$ then it would be evident that the series diverges but the alternating between high values throws me off. Any suggestions?
 A: For $|x| >1$ the sum does  not even exist so the question of its convergence as $ x \to \infty$ does not arise. 
A: The other answers have already discussed the literal meaning of the convergence of the sum, but one can ascribe what is known as an antilimit to the sum
$$\sum_{n=0}^\infty (-1)^n\frac{ x |x|^{2n}}{2n+1} \to \arctan(x)$$
This limit is exact when $|x|<1$ but what makes it an antilimit is the assignment of this value to the sum even when $|x|\geq 1$, like a principal value-esque operation. With this interpretation for the regularization of the sum, we have that
$$\lim_{x\to\infty} \arctan(x) = \frac{\pi}{2}$$
but if I was asking this question to someone, this would've been a poor way to go about it without further context, and saying the limit does not exist would be more correct.
A: We have that for $|x|>1$
$$|a_n|=\frac{\mid x\mid^{2n+1}}{2n+1}\to \infty$$
therefore the necessary condition for the convergence doesn’t hold.
A: For $x \ge 0$ we have $\sum_{n=0}^\infty (-1)^{n} \frac{x\mid x \mid^{2n}}{2n+1}=\sum_{n=0}^\infty (-1)^{n} \frac{x^{2n+1}}{2n+1}.$ This power series has radius of convergence $=1$.
Consequence: For $x>1$, the series in question is divergent.
A: To add on the previous answer, a necessary condition for the convergence of a series is that the $n^{th}$ term goes to $0$ as $n\to\infty$. In this case, if $|x|>1$, the absolute value general term is growing to $\infty$, rather than going to $0$. 
If $0\leq x\leq 1$ the series has general term of alternating sign and whose absolute value goes to zero decreasingly. This is indeed sufficient for convergence of the series.
Thus, for $x\in[0,1]$ the series converges, for $x\in(1,+\infty)$ it does not.
