# Prove by induction that for every n ≥ 1 we have?

Jul 30, 2018

$\frac{n}{2 n + 1} + \frac{1}{\left(2 \left(n + 1\right) - 1\right) \left(2 \left(n + 1\right) + 1\right)}$
=$\frac{n + 1}{2 \left(n + 1\right) + 1}$
Which checks with the series for term (n+1)

#### Explanation:

Let ${S}_{n}$ be our expression, i.e. $\frac{1}{1 \cdot 3} + \ldots + \frac{1}{\left(2 n - 1\right) \left(2 n + 1\right)}$

First we must check that it fits for n=1:

Left side: ${S}_{1} = \frac{1}{1 \cdot 3} = \frac{1}{3}$

Right side: $= \frac{1}{2 \cdot 1 + 1} = \frac{1}{3}$ Check

So if it is correct for ${S}_{n}$, we want to show that
${S}_{n + 1} = \frac{n + 1}{2 \left(n + 1\right) + 1}$

We have:
${S}_{n + 1} = {S}_{n} + \frac{1}{\left(2 \left(n + 1\right) - 1\right) \left(2 \left(n + 1\right) + 1\right)}$

$\frac{n}{2 n + 1} + \frac{1}{\left(2 \left(n + 1\right) - 1\right) \left(2 \left(n + 1\right) + 1\right)}$

=$\frac{n}{2 n + 1} + \frac{1}{\left(2 n + 1\right) \left(2 n + 3\right)}$

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

=$\frac{\left(2 n + 1\right) \left(n + 1\right)}{\left(2 n + 1\right) \left(2 n + 3\right)}$

=$\frac{n + 1}{2 n + 3} = \frac{n + 1}{2 \left(n + 1\right) + 1}$

As this is the same as the right side of our series for term (n+1), the expression checks.