The given series is:

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

To determine whether the series is convergent, divergent, oscillating, or monotonic, we analyze it as follows:

Step 1: Recognize the series structure

The series resembles the general exponential series:

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

If we take $x = 2$, the exponential series becomes:

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

Step 2: Adjust for the starting index

The given series starts from $n = 1$, whereas the exponential series starts from $n = 0$. Therefore:

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

Simplify:

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

Since $e^2$ is finite ($e \approx 2.718$), the series sum is also finite.

Step 3: Conclusion

Because the series sums to a finite value, it is convergent.

Correct answer: (A) Convergent

Do you want a detailed explanation of exponential series or any related topic?

Related questions:

1. What is the exponential series, and how does it converge?
2. How does the ratio test determine the convergence of a series?
3. Can you generalize the convergence for $\sum_{n=1}^\infty \frac{x^n}{n!}$?
4. What are the conditions for a series to be monotonic?
5. What is the difference between absolute and conditional convergence?

Tip:

The exponential series $e^x$ converges for all real values of $x$, making it a very useful tool in analyzing similar series.