The series provided in the image is:

$\sum_{n=0}^{\infty} \frac{x^{2n-1}}{n!}$

Let's simplify and solve it.

Step 1: Recognizing the series

The given series is similar to the Maclaurin series of the exponential function:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

However, we notice that the power of $x$ in our series is $2n-1$ instead of $n$. This modification affects the structure.

Step 2: Rewriting the series

We can rewrite the power $x^{2n-1}$ as:

$\sum_{n=0}^{\infty} \frac{x^{2n-1}}{n!} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$

Factoring out $x^{-1}$, we now focus on summing the series $\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$.

Step 3: Recognizing a known series

This series:

$\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$

is actually the Maclaurin series of $e^{x^2}$, because:

$e^{x^2} = \sum_{n=0}^{\infty} \frac{(x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}$

Step 4: Substituting back

Thus, the sum becomes:

$\frac{1}{x} e^{x^2}$

Final Answer:

$\sum_{n=0}^{\infty} \frac{x^{2n-1}}{n!} = \frac{e^{x^2}}{x}$

Let me know if you'd like further clarification or more details on the solution!

Here are 5 related questions to explore further:

1. How would the series change if the exponent was $2n$ instead of $2n-1$?
2. Can this method be applied to other modifications of the exponential series?
3. What happens to the sum if we replace $n!$ with $(2n)!$ in the denominator?
4. How is the solution different for series that involve odd powers of $n$?
5. How would you approach this problem if the summation index started at $n=1$ instead of $n=0$?

Tip: Understanding the Maclaurin series for common functions like $e^x$ can greatly simplify solving modified series problems like this one.