The given problem asks to determine if the series

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

is convergent or divergent.

Solution Strategy:

We can use the Ratio Test to determine the convergence or divergence of the series. The Ratio Test states that for a series $\sum a_n$, if

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L,$

• If $L < 1$, the series converges.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

Here, the general term of the series is:

$a_n = \frac{2^n (n+1)}{n!}.$

We now apply the Ratio Test.

Step 1: Find the ratio $\frac{a_{n+1}}{a_n}$.

We need to compute:

$\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1}((n+1)+1)}{(n+1)!}}{\frac{2^n (n+1)}{n!}} = \frac{2^{n+1}(n+2)}{(n+1)!} \times \frac{n!}{2^n (n+1)}.$

Simplifying:

$\frac{a_{n+1}}{a_n} = \frac{2 \cdot (n+2)}{n+1}.$

Step 2: Take the limit as $n \to \infty$.

$\lim_{n \to \infty} \frac{2 \cdot (n+2)}{n+1} = \lim_{n \to \infty} \frac{2n + 4}{n + 1} = 2.$

Step 3: Interpret the result.

Since the limit is $L = 2$, which is greater than 1, by the Ratio Test, the series diverges.

Conclusion:

The given series is divergent.

Would you like further details on any part of the solution or have additional questions?

Here are 5 related questions that might interest you:

1. How is the Ratio Test applied to other types of series?
2. What are the conditions under which the Ratio Test is inconclusive?
3. Can the Root Test be used to determine convergence for this series?
4. What are some examples of series that converge by the Ratio Test?
5. How does factorial growth affect the convergence of series?

Tip: When applying the Ratio Test, always simplify the ratio between consecutive terms before taking the limit—it helps identify divergence or convergence efficiently!