Question:

Let $a_1, a_2, \ldots$ be a geometric progression (GP) with first term $a_1$ and common ratio $r$. The sum of an infinite GP is given by $S = \sum_{k=1}^{\infty} a_k = \frac{a_1}{1-r}$, provided that $|r|<1$. From this, we can express $a_1$ in terms of $S$ and $r$: $$ a_1 = S(1-r) $$ Now, consider the series $\sum_{k=1}^{\infty} a_k^3$. The terms of this series are $a_1^3, a_2^3, a_3^3, \ldots$. Since $a_k = a_1 r^{k-1}$, we have $a_k^3 = (a_1 r^{k-1})^3 = a_1^3 (r^3)^{k-1}$. This shows that $a_k^3$ also forms a geometric progression with first term $A_1 = a_1^3$ and common ratio $R = r^3$. Since $|r|<1$, it follows that $|r^3|<1$, so the sum of this new GP also converges. The sum is given by: $$ \sum_{k=1}^{\infty} a_k^3 = \frac{A_1}{1-R} = \frac{a_1^3}{1-r^3} $$ Substitute the expression for $a_1$ from the first step into this formula: $$ \sum_{k=1}^{\infty} a_k^3 = \frac{(S(1-r))^3}{1-r^3} $$ $$ \sum_{k=1}^{\infty} a_k^3 = \frac{S^3(1-r)^3}{1-r^3} $$ Using the difference of cubes factorization $1-r^3 = (1-r)(1+r+r^2)$, we can simplify the expression: $$ \sum_{k=1}^{\infty} a_k^3 = \frac{S^3(1-r)^3}{(1-r)(1+r+r^2)} $$ $$ \sum_{k=1}^{\infty} a_k^3 = \frac{S^3(1-r)^2}{1+r+r^2} $$