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

Options

A. e
B. `e^(-1)`
C. 2e
D. None

Answer

To find the sum of the series

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

we can start by rewriting the term in the series.

### Step 1: Rewrite the term

The term \(\frac{2n}{(2n+1)!}\) can be rewritten as:

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

This allows us to express the series in a more manageable form.

### Step 2: Substitute into the series

Now we can substitute this back into the series:

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

### Step 3: Split the series

We can split this into two separate series:

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

### Step 4: Recognize the series

The first series, \(\sum_{n=0}^{\infty} \frac{1}{(2n)!}\), is known to equal \(\cosh(1)\) (the hyperbolic cosine function), and the second series, \(\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}\), is known to equal \(\sinh(1)\) (the hyperbolic sine function).

However, we need to adjust our indices since our series starts from \(n=1\):

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

### Step 5: Substitute back into the equation

Now substituting these back into our expression for \(S\):

\[ S = 2 \left( \cosh(1) - 1 \right) - 2 \sinh(1) \]

### Step 6: Simplify

Using the identities \(\cosh(1) = \frac{e + e^{-1}}{2}\) and \(\sinh(1) = \frac{e - e^{-1}}{2}\):

\[ S = 2 \left( \frac{e + e^{-1}}{2} - 1 \right) - 2 \left( \frac{e - e^{-1}}{2} \right) \]

This simplifies to:

\[ S = (e + e^{-1} - 2) - (e - e^{-1}) = 2e^{-1} \]

### Final Result

Thus, the sum of the series is:

\[ S = 2e^{-1} \]

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