Solving the Infinite Series Summation

The problem requires finding the value of the infinite series $\sum_{n=0}^{\infty} \frac{2}{(2n+1)(2n+3)}$. We can solve this using partial fraction decomposition and identifying it as a telescoping series.

Partial Fraction Decomposition

First, decompose the fraction $\frac{2}{(2n+1)(2n+3)}$ into partial fractions:

$ \frac{2}{(2n+1)(2n+3)} = \frac{A}{2n+1} + \frac{B}{2n+3} $

Multiplying both sides by $(2n+1)(2n+3)$ gives:

$ 2 = A(2n+3) + B(2n+1) $

To find A, set $n = -1/2$: $2 = A(2(-1/2)+3) + B(0) \implies 2 = A(-1+3) \implies 2 = 2A \implies A=1$.

To find B, set $n = -3/2$: $2 = A(0) + B(2(-3/2)+1) \implies 2 = B(-3+1) \implies 2 = -2B \implies B=-1$.

Thus, the expression becomes:

$ \frac{1}{2n+1} - \frac{1}{2n+3} $

Telescoping Series Calculation

Now, we evaluate the sum:

$ S = \sum_{n=0}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) $

Let's write out the first few terms of the series to see the pattern:

• For $n=0$: $\frac{1}{1} - \frac{1}{3}$
• For $n=1$: $\frac{1}{3} - \frac{1}{5}$
• For $n=2$: $\frac{1}{5} - \frac{1}{7}$
• For $n=3$: $\frac{1}{7} - \frac{1}{9}$
• ...

Consider the N-th partial sum, $S_N$:

$ S_N = \sum_{n=0}^{N} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) $

$ S_N = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots + \left(\frac{1}{2N+1} - \frac{1}{2N+3}\right) $

Most terms cancel out (telescoping effect). The remaining terms are:

$ S_N = 1 - \frac{1}{2N+3} $

Final Value Calculation

To find the value of the infinite series, we take the limit of the partial sum as $N$ approaches infinity:

$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( 1 - \frac{1}{2N+3} \right) $

As $N \to \infty$, the term $\frac{1}{2N+3}$ approaches 0.

$ S = 1 - 0 = 1 $

The value of the infinite series is 1.