Chapter 9.4, Problem 24E

To determine

To proof: The expression ${\displaystyle \sum _{i=1}^n\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{n}{2n+1}$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression ${\displaystyle \sum _{i=1}^n\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{n}{2n+1}$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider expression ${\displaystyle \sum _{i=1}^n\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{n}{2n+1}$ .

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

  $\begin{array}{c}\frac{1}{\left(2\left(1\right)-1\right)\left(2\left(1\right)+1\right)}=\frac{1}{2\left(1\right)+1}\\ \frac{1}{1\left(3\right)}=\frac{1}{3}\\ \frac{1}{3}=\frac{1}{3}\end{array}$

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ .

  ${\displaystyle \sum _{i=1}^k\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{k}{2k+1}$  ...(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ . Group the terms that involve $S_k$ .

  ${\displaystyle \sum _{i=1}^{k+1}\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=S_k+\frac{k+1}{2\left(k+1\right)+1}$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=S_k+\frac{1}{\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)}\\ =\frac{k}{2k+1}+\frac{1}{\left(2\left(k+1\right)-1\right)\left(2\left(k+1\right)+1\right)}\\ =\frac{k}{2k+1}+\frac{1}{\left(2k+1\right)\left(2k+3\right)}\end{array}$

Take the common terms and simplify it further as,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{1}{\left(2k+1\right)}\left[k+\frac{1}{2k+3}\right]\\ =\frac{2k^2+3k+1}{\left(2k+1\right)\left(2k+3\right)}\\ =\frac{2k^2+2k+k+1}{\left(2k+1\right)\left(2k+3\right)}\\ =\frac{2k\left(k+1\right)+1\left(k+1\right)}{\left(2k+1\right)\left(2k+3\right)}\end{array}$

Further simplify it as,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{2k\left(k+1\right)+1\left(k+1\right)}{\left(2k+1\right)\left(2k+3\right)}\\ =\frac{\left(2k+1\right)\left(k+1\right)}{\left(2k+1\right)\left(2k+3\right)}\\ =\frac{k+1}{2k+3}\\ =\frac{k+1}{2\left(k+1\right)+1}\end{array}$

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  ${\displaystyle \sum _{i=1}^{k+1}\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{k+1}{2\left(k+1\right)+1}$

Hence, by the principle of mathematical induction the statement ${\displaystyle \sum _{i=1}^n\frac{1}{\left(2i-1\right)\left(2i+1\right)}}=\frac{n}{2n+1}$ is true for all integers $n\ge 1$ .