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$$\lim_{x \rightarrow \infty} \sum_{n=0}^\infty (-1)^{n} \frac{x \lvert x \rvert^{2n}}{2n+1}$$ 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?

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    $\begingroup$ @ 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 $\frac{\pi}{2}$ $\endgroup$ – Dr. Wolfgang Hintze Dec 6 '19 at 10:05
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For $|x| >1$ the sum does not even exist so the question of its convergence as $ x \to \infty$ does not arise.

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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.

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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.

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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.

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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.

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