`sum_(n=1)^(oo) (1)/(2n(2n+1))` is equal to

Options

A. `log_(e )(2/e)`
B. `1-log_(e)2`
C. `1-log_(e)(1)/(2e)`
D. `log(1)/(2e)`

Explanation

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

### Step 1: Rewrite the term
We start with the term \( \frac{1}{2n(2n+1)} \). We can simplify this using partial fractions:
\[ \frac{1}{2n(2n+1)} = \frac{A}{2n} + \frac{B}{2n+1} \]
Multiplying through by the common denominator \( 2n(2n+1) \) gives:
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