Question

Show if nis a positive integer with \(n \ge 2\), then

\(\sum\limits_{j = 2}^n {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {n - 1} \right)\left( {3n + 2} \right)}}{{4n\left( {n + 1} \right)}}\)

Short answer

It is shown that\(\sum\limits_{j = 2}^n {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {n - 1} \right)\left( {3n + 2} \right)}}{{4n\left( {n + 1} \right)}}\).

Step-by-step solution

Principle of Mathematical Induction

To prove that\(P\left( n \right)\)is true for all positive integers n, where\(P\left( n \right)\)is a propositional function, it completes two steps:

Basis Step:

It verifies that\(P\left( 1 \right)\)is true.

Inductive Step:

It shows that the conditional statement \(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

Proving the basis step

Let\(P\left( n \right)\): “\(\sum\limits_{j = 2}^n {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {n - 1} \right)\left( {3n + 2} \right)}}{{4n\left( {n + 1} \right)}}\)”.

In the basis step, it needs to prove that\(P\left( 1 \right)\)is true.

Since\(n \ge 2\), it needs to prove that\(P\left( 2 \right)\)is true.

For finding statement\(P\left( 2 \right)\)substituting 2 for\(n\)in the statement:

\(\begin{array}{c}\frac{1}{{{{\left( 2 \right)}^2} - 1}} = \frac{{\left( {2 - 1} \right)\left( {3\left( 2 \right) + 2} \right)}}{{4\left( 2 \right)\left( {2 + 1} \right)}}\\\frac{1}{{4 - 1}} = \frac{8}{{8 \cdot 3}}\\\frac{1}{3} = \frac{1}{3}\end{array}\)

From the above, it can see that the statement \(P\left( 2 \right)\) is true this is also known as the basis step of the proof.

Proving the Inductive step

In the inductive step, it needs to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, it assumes that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).

That is\(\sum\limits_{j = 2}^k {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {k - 1} \right)\left( {3k + 2} \right)}}{{4k\left( {k + 1} \right)}}\).

Now it must have to show that\(P\left( {k + 1} \right)\)is also true.

Therefore, replacing\(k\)with\(k + 1\)in the statement:

From inductive hypothesis it knows\(\sum\limits_{j = 2}^k {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {k - 1} \right)\left( {3k + 2} \right)}}{{4k\left( {k + 1} \right)}}\).

Therefore, it is written as:

\(\begin{aligned}{c}\sum\limits_{j = 2}^{k + 1} {\frac{1}{{{j^2} - 1}}} &= \frac{{\left( {k - 1} \right)\left( {3k + 2} \right)}}{{4k\left( {k + 1} \right)}} + \frac{1}{{{{\left( {k + 1} \right)}^2} - 1}}\\ &= \frac{{\left( {k - 1} \right)\left( {3k + 2} \right)}}{{4k\left( {k + 1} \right)}} + \frac{1}{{{k^2} + 2k}}\\ &= \frac{{\left( {k - 1} \right)\left( {3k + 2} \right)}}{{4k\left( {k + 1} \right)}} + \frac{1}{{k\left( {k + 2} \right)}}\\ &= \frac{{\left( {k + 2} \right)\left( {k - 1} \right)\left( {3k + 2} \right) + 4\left( {k + 1} \right)}}{{4k\left( {k + 1} \right)\left( {k + 2} \right)}}\end{aligned}\)

Further, solve the above expression,

\(\begin{aligned}{c}\sum\limits_{j = 2}^{k + 1} {\frac{1}{{{j^2} - 1}}} &= \frac{{\left( {k + 2} \right)\left( {k - 1} \right)\left( {3k + 2} \right) + 4\left( {k + 1} \right)}}{{4k\left( {k + 1} \right)\left( {k + 2} \right)}}\\ &= \frac{{3{k^3} + 5{k^2}}}{{4k\left( {k + 1} \right)\left( {k + 2} \right)}}\end{aligned}\)

Canceling similar terms,

\(\begin{aligned}{c}\sum\limits_{j = 2}^{k + 1} {\frac{1}{{{j^2} - 1}}} &= \frac{{3{k^2} + 5k}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\\ &= \frac{{k\left( {3k + 5} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\\ &= \frac{{\left( {\left( {k + 1} \right) - 1} \right)\left( {3\left( {k + 1} \right) + 2} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}\end{aligned}\)

From the above, it can see that\(P\left( {k + 1} \right)\)is also true.

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This

completes the inductive step.

Hence It is shown that\(\sum\limits_{j = 2}^n {\frac{1}{{{j^2} - 1}}} = \frac{{\left( {n - 1} \right)\left( {3n + 2} \right)}}{{4n\left( {n + 1} \right)}}\).