Question

If the sum of the n terms of the series \[\dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + ...\] is \[{S_n}\], then \[{S_n}\] exceeds \[1.99\] for all n greater than.

Answer 1

Hint: At first, we will consider the general term as, \[{T_n} = \dfrac{{1 + 2 + 3 + ... + n}}{{{1^3} + {2^3} + {3^3} + ... + {n^3}}}\]
According to the problem we have then,
\[{S_n} = \sum\limits_{n = 1}^n {{T_n}} \]
The, we will apply two formulas:
The sum of first n natural numbers that is \[1 + 2 + 3 + ... + n = \dfrac{{n(n + 1)}}{2}\]
The sum of the cube of first n natural numbers that is \[{1^3} + {2^3} + {3^3} + ... + {n^3} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\]
By solving, we will get our required answer that is the value of \[n\] for which \[{S_n}\] exceeds \[1.99\].

Complete step-by-step solution:
It is given that: the sum of the n terms of the series \[\dfrac{1}{{{1^3}}} + \dfrac{{1 + 2}}{{{1^3} + {2^3}}} + \dfrac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + ...\] is \[{S_n}\]. And \[{S_n}\] exceeds \[1.99\].
We have to find the value of \[n\] for which \[{S_n}\] exceeds \[1.99\].
Let us consider,
\[{T_n} = \dfrac{{1 + 2 + 3 + ... + n}}{{{1^3} + {2^3} + {3^3} + ... + {n^3}}}\]
So, as per the problem,
\[{S_n} = \sum\limits_{n = 1}^n {{T_n}} \]
Substitute the value we get,
$\Rightarrow$\[{S_n} = \sum\limits_{n = 1}^n {\dfrac{{1 + 2 + 3 + ... + n}}{{{1^3} + {2^3} + {3^3} + ... + {n^3}}}} \]
We know that,
The sum of first n natural numbers that is \[1 + 2 + 3 + ... + n = \dfrac{{n(n + 1)}}{2}\]
The sum of the cube of first n natural numbers that is \[{1^3} + {2^3} + {3^3} + ... + {n^3} = {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\]
Substitute the values we get,
$\Rightarrow$\[{S_n} = \sum\limits_{n = 1}^n {\dfrac{{\dfrac{{n(n + 1)}}{2}}}{{{{\left[ {\dfrac{{n(n + 1)}}{2}} \right]}^2}}}} \]
Simplifying we get,
$\Rightarrow$\[{S_n} = \sum\limits_{n = 1}^n {\dfrac{2}{{n{{(n + 1)}^2}}}} \]
Simplifying again we get,
$\Rightarrow$\[{S_n} = 2\sum\limits_{n = 1}^n {\left[ {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right]} \]
Expanding the terms, we get,
$\Rightarrow$\[{S_n} = 2\left[ {1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... - \dfrac{1}{n} + \dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right]\]
Simplifying we get,
$\Rightarrow$\[{S_n} = 2\left[ {1 - \dfrac{1}{{n + 1}}} \right]\]
Simplifying again we get,
$\Rightarrow$\[{S_n} = \dfrac{{2n}}{{n + 1}}\]
If let \[n < 2 \Rightarrow n = 1\],
$\Rightarrow$\[n = 1 \Rightarrow {S_1} = \dfrac{{2\left( 1 \right)}}{{1 + 1}} = \dfrac{2}{2} = 1\]
If \[n = 2 \Rightarrow {S_2} = \dfrac{{2\left( 2 \right)}}{{2 + 1}} = \dfrac{4}{3} = 1.33...\]

\[{S_n}\] exceeds \[1.99\] for all n greater than when \[n < 2\].

Note: We have to remember that, an itemized collection of elements in which repetitions of any sort are allowed is known as a sequence. A series can be highly generalized as the sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence.