Question

How do you prove that the sum of infinite series $1 + \dfrac{1}{4} + \dfrac{1}{9} + ............$ is less than two?

Answer 1

Hint: In this question we will write the infinite series in the form of a summation which ranges from a variable $n$ which takes the value from $0$ to $\infty $ and then use the properties of summation to simplify the expression and prove the statement.

Complete step-by-step solution:
We have the infinite series as:
$ \Rightarrow 1 + \dfrac{1}{4} + \dfrac{1}{9} + ............$
It can be written in the form of summation as:
$ \Rightarrow \sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^2}}}} $
Now this expression can be split up as:
$ \Rightarrow \sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^2}}}} = 1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{{n^2}}}} $
Now if we subtract the number $n$from the denominator, the value of the infinite series will be lesser than of series $1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{{n^2}}}} $therefore, we can write it as:
$ \Rightarrow 1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{{n^2}}}} < 1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{{n^2} - n}}} $
Now on taking the common term from the denominator, we get:
$ \Rightarrow 1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{{n^2}}}} < 1 + \sum\limits_{n = 2}^\infty {\dfrac{1}{{n(n - 1)}}} $
Now let’s consider the term $\dfrac{1}{{n(n - 1)}}$.
Now we know that if we add and subtract a value in an expression at the same time, the value of the expression does not change therefore, we will add and subtract $n$in the numerator. We can write it as:
$ \Rightarrow \dfrac{{n - n + 1}}{{n(n - 1)}}$
Now the terms can be grouped and written as:
$ \Rightarrow \dfrac{{n - (n - 1)}}{{n(n - 1)}}$
Now on splitting the fraction, we can write it as:
$ \Rightarrow \dfrac{n}{{n(n - 1)}} - \dfrac{{(n - 1)}}{{n(n - 1)}}$
Now on simplifying, we get:
$ \Rightarrow \dfrac{1}{{(n - 1)}} - \dfrac{1}{n}$
Therefore, we can write the summation $\sum\limits_{n = 2}^\infty {\dfrac{1}{{n(n - 1)}}} $ as:
\[ \Rightarrow \sum\limits_{n = 2}^\infty {\dfrac{1}{{(n - 1)}} - \dfrac{1}{n}} \]
Now on splitting the summation, we get:
\[ \Rightarrow \sum\limits_{n = 2}^\infty {\dfrac{1}{{(n - 1)}} - \sum\limits_{n = 2}^\infty {\dfrac{1}{n}} } \]
Now on splitting the summation as a sum of $1$, we get:
\[ \Rightarrow 1 + \sum\limits_{n = 3}^\infty {\dfrac{1}{{(n - 1)}} - \sum\limits_{n = 2}^\infty {\dfrac{1}{n}} } \]
Now on subtracting a term $N$ from the summation, we get:
\[ \Rightarrow 1 + \sum\limits_{n = 3}^{\infty - 1} {\dfrac{1}{{(n - 1)}} - \sum\limits_{n = 2}^{\infty - 1} {\dfrac{1}{n}} - \dfrac{1}{N}} \]
Now since both the summations are same, they can be cancelled and written as:
$ \Rightarrow 1 - \dfrac{1}{N}$
So, we can deduce that:
$ \Rightarrow \sum\limits_{n = 2}^\infty {\dfrac{1}{{n(n - 1)}}} = \mathop {\lim }\limits_{N1 \to \infty } \left( {1 - \dfrac{1}{N}} \right) = 1$
Therefore, the value of $\sum\limits_{n = 2}^\infty {\dfrac{1}{{n(n - 1)}}} $can be written as $1 + 1 = 2$, which is not the sum of the expression $ \Rightarrow \sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^2}}}} $

Therefore, the value of $\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^2}}}} < 2$, hence proved.

Note: The properties of summation and limits should be remembered while doing these types of sums.
It is to be remembered that an infinite series is a sum of infinite terms which follow a certain rule for every iteration in the term.
It can be written in a simplified manner using the summation property.