# 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?

• @ hwood87 Taken plainly the sum is divergent unless $|x|<1$. However, if understood as the limit of the analytic continuation of the sum $\frac{x \operatorname{arctan}(\left| x\right| )}{\left| x\right| }$ to $|x|$ >1 then the limit is $\frac{\pi}{2}$ – Dr. Wolfgang Hintze Dec 6 '19 at 10:05

For $$|x| >1$$ the sum does not even exist so the question of its convergence as $$x \to \infty$$ does not arise.

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.

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.

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.

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.