x/1+x + x^2/1+x^2 + x^3/1+x^3... 
Check converge x=-1

To check the convergence of the series 

\[
\frac{x}{1+x} + \frac{x^2}{1+x^2} + \frac{x^3}{1+x^3} + \cdots 
\]

at \(x = -1\), we need to evaluate each term in the series at \(x = -1\) and then determine if the series converges or diverges.

Let's evaluate each term at \(x = -1\):

1. **First term**: \( \frac{x}{1+x} \)  
   \[
   \frac{-1}{1 + (-1)} = \frac{-1}{0}.
   \]

The first term is undefined because it involves division by zero.

Since the series contains an undefined term at \(x = -1\), the series does not converge at \(x = -1\). It is instead undefined at this point.

### Conclusion:
The series **does not converge** at \(x = -1\) due to the presence of a division by zero in the first term.

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Would you like more details on this, or do you have any questions?

Here are some related questions to consider:
1. What happens if \(x\) is another value, such as \(x = 0.5\) or \(x = 2\)?
2. How do you determine the radius of convergence for a power series?
3. What are other methods to check the convergence of a series?
4. What does it mean for a series to converge conditionally versus absolutely?
5. How does the behavior of a series change at its endpoints of convergence?

**Tip:** When checking convergence, always carefully examine if any term could lead to undefined expressions or infinite values.

Explore how to check the convergence of the series x/(1+x) + x^2/(1+x^2) + x^3/(1+x^3)... at x=-1. Understand why the series does not converge at x=-1 due to division by zero in the first term. Learn about series convergence and limit theorems in calculus.

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