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

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.