Key Concepts and Formulas

• Sum of the first $n$ odd natural numbers: The sum $1 + 3 + 5 + \ldots + (2n-1)$ is equal to $n^2$.
• Series expansion of $e$: The Maclaurin series for $e$ is given by $e = \sum_{n=0}^{\infty} \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots$.
• Properties of Factorials: $n! = n \times (n-1)!$ for $n \ge 1$, and $0! = 1$.

Step-by-Step Solution

Step 1: Identify the general term of the given series. The given series is $S = 1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$. Let $T_n$ denote the $n$-th term of the series. The denominator of the $n$-th term is $n!$. The numerator of the $n$-th term is the sum of the first $n$ odd natural numbers: $1+3+5+\ldots+(2n-1)$. Using the formula for the sum of the first $n$ odd natural numbers, the numerator is $n^2$. Therefore, the general term is $T_n = \frac{n^2}{n!}$ for $n \ge 1$.

Step 2: Simplify the general term $T_n$. We can simplify $T_n = \frac{n^2}{n!}$ by using the property $n! = n \times (n-1)!$: $T_n = \frac{n^2}{n \times (n-1)!} = \frac{n}{(n-1)!}$ This simplified form is valid for $n \ge 1$.

Step 3: Rewrite the simplified general term $T_n$ to relate it to the series expansion of $e$. We can rewrite the numerator $n$ as $(n-1) + 1$: $T_n = \frac{(n-1) + 1}{(n-1)!}$ Now, we can split this into two fractions: $T_n = \frac{n-1}{(n-1)!} + \frac{1}{(n-1)!}$ For $n \ge 2$, we can further simplify the first term using $(n-1)! = (n-1) \times (n-2)!$: $T_n = \frac{n-1}{(n-1)(n-2)!} + \frac{1}{(n-1)!} = \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$ This form is valid for $n \ge 2$.

Step 4: Express the sum of the series using the simplified general term. The sum of the series is $S = \sum_{n=1}^{\infty} T_n$. We can write the sum by separating the first term ($n=1$) from the rest of the terms ($n \ge 2$): $S = T_1 + \sum_{n=2}^{\infty} T_n$ From Step 1, $T_1 = \frac{1^2}{1!} = 1$. Using the simplified form from Step 3 for $n \ge 2$: $S = 1 + \sum_{n=2}^{\infty} \left( \frac{1}{(n-2)!} + \frac{1}{(n-1)!} \right)$ We can split the summation into two separate sums: $S = 1 + \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=2}^{\infty} \frac{1}{(n-1)!}$

Step 5: Evaluate the summations. For the first summation, $\sum_{n=2}^{\infty} \frac{1}{(n-2)!}$: Let $k = n-2$. When $n=2$, $k=0$. As $n \to \infty$, $k \to \infty$. So, $\sum_{n=2}^{\infty} \frac{1}{(n-2)!} = \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots = e$.

For the second summation, $\sum_{n=2}^{\infty} \frac{1}{(n-1)!}$: Let $m = n-1$. When $n=2$, $m=1$. As $n \to \infty$, $m \to \infty$. So, $\sum_{n=2}^{\infty} \frac{1}{(n-1)!} = \sum_{m=1}^{\infty} \frac{1}{m!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots$. We know that $e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \ldots = 1 + \left( \frac{1}{1!} + \frac{1}{2!} + \ldots \right)$. Therefore, $\sum_{m=1}^{\infty} \frac{1}{m!} = e - 1$.

Step 6: Combine the results to find the total sum. Substitute the values of the summations back into the expression for $S$: $S = 1 + (e) + (e-1)$ $S = 1 + e + e - 1$ $S = 2e$

Alternative Method: Using $T_n = \frac{n}{(n-1)!}$ directly

We can also evaluate the sum $S = \sum_{n=1}^{\infty} \frac{n}{(n-1)!}$ by making a substitution. Let $k = n-1$. Then $n = k+1$. When $n=1$, $k=0$. $S = \sum_{k=0}^{\infty} \frac{k+1}{k!}$ Split the fraction: $S = \sum_{k=0}^{\infty} \left( \frac{k}{k!} + \frac{1}{k!} \right)$ $S = \sum_{k=0}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!}$ The second sum is $\sum_{k=0}^{\infty} \frac{1}{k!} = e$. For the first sum, $\sum_{k=0}^{\infty} \frac{k}{k!}$, the term for $k=0$ is $\frac{0}{0!} = 0$. For $k \ge 1$, $\frac{k}{k!} = \frac{k}{k(k-1)!} = \frac{1}{(k-1)!}$. So, $\sum_{k=0}^{\infty} \frac{k}{k!} = 0 + \sum_{k=1}^{\infty} \frac{1}{(k-1)!}$. Let $j = k-1$. When $k=1$, $j=0$. $\sum_{k=1}^{\infty} \frac{1}{(k-1)!} = \sum_{j=0}^{\infty} \frac{1}{j!} = e$. Therefore, $S = e + e = 2e$.

Common Mistakes & Tips

• Handling the first term: Be careful when simplifying the general term. The simplification $\frac{1}{(n-2)!}$ is not valid for $n=1$, so it's often best to separate the first term of the series.
• Index of summation: When changing variables in summations (e.g., $k=n-2$), ensure the new limits of summation are correctly determined.
• Recognizing the series for $e$: The ability to recognize and manipulate series that sum to $e$ is crucial for solving this type of problem.

Summary

The problem requires finding the sum of an infinite series. We first identified the general term of the series as $T_n = \frac{n^2}{n!}$. By simplifying this term to $T_n = \frac{1}{(n-2)!} + \frac{1}{(n-1)!}$ for $n \ge 2$, and handling the first term separately, we were able to express the sum as a combination of series that sum to $e$. Evaluating these series yielded a total sum of $2e$. An alternative method by rewriting $T_n = \frac{n}{(n-1)!}$ and making a substitution also leads to the same result.

The final answer is $\boxed{2e}$.