The image shows the sum:

$\sum_{n=1}^{\infty} \frac{1}{(2n+1)(2n+3)}$

This is an infinite series that involves rational functions. To solve it, we can attempt partial fraction decomposition.

We begin by decomposing the general term into partial fractions:

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

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

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

Now, expand the terms:

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

For this to hold true for all $n$, we need to match the coefficients:

1. $A + B = 0$
2. $3A + B = 1$

Solving these equations:

From $A + B = 0$, we get $B = -A$.

Substitute $B = -A$ into $3A + B = 1$:

$3A - A = 1 \quad \Rightarrow \quad 2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}$

Thus, $B = -\frac{1}{2}$.

Now, we rewrite the general term:

$\frac{1}{(2n+1)(2n+3)} = \frac{1/2}{2n+1} - \frac{1/2}{2n+3}$

Therefore, the series becomes:

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

This is a telescoping series, meaning most terms cancel out. Writing out the first few terms:

$\frac{1/2}{3} - \frac{1/2}{5} + \frac{1/2}{5} - \frac{1/2}{7} + \frac{1/2}{7} - \frac{1/2}{9} + \cdots$

The result is that all terms cancel except for the first term, $\frac{1/2}{3}$. So, the sum of the series is:

$\frac{1}{6}$

Let me know if you'd like more details or have any questions about this solution!

Related Questions:

1. How can we determine if a series converges before solving it?
2. What is the general method for summing a telescoping series?
3. Can this method be applied to other infinite series?
4. What is partial fraction decomposition and why does it work?
5. How do you calculate the sum of a series with alternating terms?

Tip: When solving series like this, look for patterns like telescoping behavior or use partial fraction decomposition to simplify complex terms.