If `a = Sigma_(n=1)^(oo) (2n)/(2n-1!),b=Sigma__(n=1)^(oo) (2n)/(2n+1!)` then ab equals

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 product \( ab \) where:

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

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

We start with the series for \( a \):

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

We can express \( 2n \) as \( 2n = (2n-1) + 1 \):

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

This separates into two sums:

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

The first sum can be simplified:

\[ \sum_{n=1}^{\infty} \frac{(2n-1)}{(2n-1)!} = \sum_{k=0}^{\infty} \frac{1}{k!} = e \]

The second sum:

\[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \ldots = \sinh(1) \]

Thus, we can express \( a \) as:

\[ a = e + \sinh(1) \]

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

Now we calculate \( b \):

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

We can express \( 2n \) as \( 2n = (2n+1) - 1 \):

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

This separates into two sums:

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

The first sum can be simplified:

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

The second sum is:

\[ \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \sinh(1) \]

Thus, we can express \( b \) as:

\[ b = \cosh(1) - \sinh(1) \]

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

Now we can find \( ab \):

\[ ab = (e + \sinh(1))(\cosh(1) - \sinh(1)) \]

Using the identity \( \cosh(1) - \sinh(1) = e^{-1} \):

\[ ab = (e + \sinh(1)) e^{-1} \]

This simplifies to:

\[ ab = 1 \]

### Conclusion

Thus, the final answer is:

\[ ab = 1 \]