`sum_(n=1)^(oo) (2n)/(n!)=`

Options

A. e
B. 2e
C. 3e
D. `1/e`

Answer

To solve the problem \( \sum_{n=1}^{\infty} \frac{2n}{n!} \), we can follow these steps:

### Step 1: Rewrite the Summation
We start with the given summation:
\[ \sum_{n=1}^{\infty} \frac{2n}{n!} \]
We can factor out the constant \(2\):
\[ = 2 \sum_{n=1}^{\infty} \frac{n}{n!} \]

### Step 2: Simplify the Term \( \frac{n}{n!} \)
Recall that \( n! = n \cdot (n-1)! \). Therefore, we can rewrite \( \frac{n}{n!} \) as:
\[ \frac{n}{n!} = \frac{1}{(n-1)!} \]
Thus, we can substitute this back into our summation:
\[ = 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} \]

### Step 3: Change the Index of Summation
To simplify the summation further, we can change the index of summation. Let \( m = n - 1 \). When \( n = 1 \), \( m = 0 \), and as \( n \) approaches infinity, \( m \) also approaches infinity. Therefore, we have:
\[ = 2 \sum_{m=0}^{\infty} \frac{1}{m!} \]

### Step 4: Recognize the Series
The series \( \sum_{m=0}^{\infty} \frac{1}{m!} \) is known to converge to \( e \) (the base of natural logarithms):
\[ = 2e \]

### Final Answer
Thus, the value of the original summation is:
\[ \sum_{n=1}^{\infty} \frac{2n}{n!} = 2e \]

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