Let the sum of the series $$\frac{1}{1^3} + \frac{1+2}{1^3+2^3} +\frac{1+2+3}{1^3+2^3+3^3} + \cdots + \dfrac{1+2+\cdots+n}{1^3+2^3+\cdots+n^3}$$ be $S_n$. Find the value of $S_n$.
I want to learn how to solve this problem using $V_n$ method. And do I have to seperate the numerator and denominator into different series?. Please help
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notice that what you have here is: $$S_n=\sum_{i=1}^n\frac{\sum_{k=1}^ik}{\sum_{j=1}^ij^3}$$ now, it is well-known that: $$\sum_{k=1}^ik=\frac{i(i+1)}{2}$$ $$\sum_{j=1}^ij^3=\frac{i^2(i+1)^2}{4}$$ and so we can see that: $$\frac{\sum_{k=1}^ik}{\sum_{j=1}^ij^3}=\frac{2}{i(i+1)}$$ and so we can now say: $$S_n=\sum_{i=1}^n\frac{2}{i(i+1)}=2\sum_{i=1}^n\left(\frac1i-\frac1{i+1}\right)$$ Now notice that this is a telescoping series
First, we can apply Power Sum formula or Faulhaber's formula to rewrite sum of cubes in the form $$ 1^3+2^3+\ldots+k^3=(1+2+\ldots+k)^2=\left(\dfrac{k(k+1)}{2}\right)^2. $$ Then the term of the series will have easier form: $$ \dfrac{1+2+\ldots+k}{1^3+2^3+\ldots+k^3}=\dfrac{1}{1+2+\ldots+k}=\dfrac{2}{k(k+1)}. $$
When we rewrite the fraction $\dfrac{2}{k(k+1)}$ as difference of fractions: $$ \dfrac{2}{k(k+1)} = \dfrac{2}{k} - \dfrac{2}{k+1}, $$ we can apply Telescoping series to find final answer.
Series then will get the form: \begin{array}{ccccccccc} \left(\dfrac{2}{1}\right. & - \left.\dfrac{2}{2}\right) + & & & & & \\ & \left(\dfrac{2}{2} \right.& - \left.\dfrac{2}{3}\right) + & & & & \\ & & \left(\dfrac{2}{3} \right.&- \left.\dfrac{2}{4}\right) + & & &\\ & & & \ldots + & & \\ & & & & \left(\dfrac{2}{k-1} \right.& - \left.\dfrac{2}{k}\right) + & \\ & & & & & \left(\dfrac{2}{k} \right.& - \left.\dfrac{2}{k+1}\right) + \\ & & & & & & \ldots + \\ & & & & & & & \left(\dfrac{2}{n} \right.& - \left.\dfrac{2}{n+1}\right) . \end{array}