If `A = sum_(n=1)^(oo) (2n)/((2n-1)!), B = sum_(n=1)^(oo) (2n)/((2n+1)!)`, then AB is equal to

Options

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

Answer

To solve the problem, we need to find the values of \( A \) and \( B \) and then compute the product \( AB \).

### Step 1: Calculate \( A \)

We start with the expression for \( A \):

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

To simplify this, we can rewrite \( 2n \) as \( (2n-1) + 1 \):

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

The first term simplifies as follows:

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

This can be rewritten by changing the index:

\[ \sum_{m=0}^{\infty} \frac{1}{(2m)!} = \cosh(1) \]

The second term is:

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

Thus, we have:

\[ A = \cosh(1) + \sinh(1) = e \]

### Step 2: Calculate \( B \)

Now, we compute \( B \):

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

Using a similar approach, we can rewrite \( 2n \) in terms of \( (2n+1) \):

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

The first term simplifies as follows:

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

This can be rewritten by changing the index:

\[ \sum_{m=0}^{\infty} \frac{1}{(2m)!} = \cosh(1) \]

The second term is:

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

Thus, we have:

\[ B = \cosh(1) - (\sinh(1) - 1) = e^{-1} \]

### Step 3: Calculate \( AB \)

Now we can find \( AB \):

\[ AB = A \cdot B = e \cdot e^{-1} = 1 \]

### Final Answer

Thus, the product \( AB \) is equal to \( 1 \).

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