Question

Let P(n) be the statement that$1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{n^2}<2-\frac{1}{n}$ , where n is an integer greater than 1.

a) What is the statement P(2)?

b) Show that P(2) is true, completing the basis step of the proof.

c) What is the inductive hypothesis?

d) What do you need to prove in the inductive step?

e) Complete the inductive step.

f) Explain why these steps show that this inequality is true whenever n is an integer greater than 1.

Short answer

a)$\begin{array}{r}P\left(2\right)=1+\frac{1}{2^2}<2-\frac{1}{2}\\ 1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{n^2}<2-\frac{1}{n}\\ n=2\end{array}$

b)$\begin{array}{r}1+\frac{1}{2^2}<2-\frac{1}{2}\\ \frac{5}{4}<\frac{3}{2}\end{array}$

Hence verified.

c) $1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{k^2}<2-\frac{1}{k}$for any k belong to natural numbers.

d) We need to prove that P(k+1) is true which says that

$1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{k^2}+\frac{1}{k+1^2}<2-\frac{1}{k+1}$

where k+1 is a positive integer greater.

e) P (k+1) is also true.

f) P (n): $1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{n^2}<2-\frac{1}{n}$is true for every natural number n greater than 1.

Step-by-step solution

Step: 1

a)$P\left(2\right)=1+\frac{1}{2^2}<2-\frac{1}{2}\mid$

b)$\begin{array}{r}1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{n^2}<2-\frac{1}{n}\\ n=2\\ 1+\frac{1}{2^2}<2-\frac{1}{2}\\ \frac{5}{4}<\frac{3}{2}\end{array}$

Hence verified.

Step: 2

c) Let P(k) be true.

That is$1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{k^2}<2-\frac{1}{k}$ for any k belong to natural numbers.

d) We need to prove that P(k+1) is true which says that

$1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{k^2}+\frac{1}{k+1^2}<2-\frac{1}{k+1}$

where k+1 is a positive integer greater.

Step: 3

1. We know,

$\begin{array}{r}1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{k^2}+\frac{1}{(k+1{)}^2}\\ <2-\frac{1}{k}+\frac{1}{(k+1{)}^2}\\ =2-\frac{k^2+k+1}{k(k+1{)}^2}\\ =2-\frac{k(k+1)}{k(k+1{)}^2}-\frac{1}{k(k+1{)}^2}\\ =2-\frac{1}{k(k+1{)}^2}-\frac{1}{k(k+1{)}^2}\\ <2-\frac{1}{k+1}\end{array}$

Hence P (k+1) is also true.

f) P(n): $1+\frac{1}{4}+\frac{1}{9}+\dots +\frac{1}{n^2}<2-\frac{1}{n}$is true for every natural number n greater than 1.