Step 1: Understanding the Concept:
This is an infinite series related to the exponential series \( e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \). For \( x=1 \), \( e = \sum_{n=0}^{\infty} \frac{1}{n!} \).
: Key Formula or Approach:
General term \( T_{n} = \frac{2n-1}{(n-1)!} \) where \( n=1, 2, 3 \dots \).
Or more simply, let the sum be \( S = \sum_{n=0}^{\infty} \frac{2n+1}{n!} \).
Step 2: Detailed Explanation:
The given series can be written as:
\[ S = \sum_{n=0}^{\infty} \frac{2n + 1}{n!} \]
Splitting the summation:
\[ S = \sum_{n=0}^{\infty} \frac{2n}{n!} + \sum_{n=0}^{\infty} \frac{1}{n!} \]
In the first part, the term for \( n=0 \) is 0. So, we start from \( n=1 \):
\[ S = 2 \sum_{n=1}^{\infty} \frac{n}{n(n-1)!} + \sum_{n=0}^{\infty} \frac{1}{n!} \]
\[ S = 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} + \sum_{n=0}^{\infty} \frac{1}{n!} \]
Both summations are equal to the series for \( e \):
\[ S = 2e + e = 3e \].
Step 3: Final Answer:
The sum of the series is 3e.