https://file.aiquickdraw.com/mathbot/file/1729802436927sd6zfa8c.jpg

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.

Find the sum of the series: ∑(n=0 to ∞) x^(2n-1) / n!

Discover how to find the sum of the series ∑(n=0 to ∞) x^(2n-1) / n! by recognizing the Maclaurin series of exponential functions. This step-by-step solution explains the transformation of the series into e^(x^2)/x. Ideal for advanced high school and college-level math.

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