The sum of the series `sum_(n=1)^(oo)(2n)/((2n+1)!)` is

Options

A. e
B. `e^(-1)`
C. 2e
D. `2e^(-1)`

Answer

To find the sum of the series

\[ S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]

we can follow these steps:

### Step 1: Rewrite the series
We start by rewriting the series in a more manageable form. The term \(\frac{2n}{(2n+1)!}\) can be expressed as:

\[ \frac{2n}{(2n+1)!} = \frac{2n}{(2n+1)(2n)!} = \frac{2}{(2n+1)!} \cdot n \]

### Step 2: Identify the pattern
Next, we can express \(n\) in terms of factorials:

\[ n = \frac{(2n)!}{(2n-1)!} = \frac{(2n)!}{(2n-1)! \cdot 2} \]

Thus, we can rewrite the series as:

\[ S = \sum_{n=1}^{\infty} \frac{2}{(2n+1)!} \cdot \frac{(2n)!}{(2n-1)! \cdot 2} \]

### Step 3: Simplify the series
This simplifies to:

\[ S = \sum_{n=1}^{\infty} \frac{(2n)!}{(2n-1)! \cdot (2n+1)!} \]

### Step 4: Recognize the series
Now we can recognize that this series resembles the Taylor series expansion for \(e^x\). Recall that:

\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]

### Step 5: Use the exponential function
To find the sum of our series, we can relate it to the exponential function. Specifically, we can use the fact that:

\[ e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \]

### Step 6: Relate to known series
By substituting \(x = -1\) into the expansion of \(e^x\), we can find:

\[ e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots \]

### Step 7: Final sum
The series converges to:

\[ S = e^{-1} = \frac{1}{e} \]

Thus, the sum of the series

\[ \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} = \frac{1}{e} \]