Consecutively different

Algebra Level 5

S 1 = 1 + 1 4 + 1 × 4 4 × 8 + 1 × 4 × 7 4 × 8 × 12 + 1 × 4 × 7 × 10 4 × 8 × 12 × 16 + S 2 = 1 + 2 6 + 2 × 5 6 × 12 + 2 × 5 × 8 6 × 12 × 18 + 2 × 5 × 8 × 11 6 × 12 × 18 × 24 + \begin{aligned} S_1 &=& 1 + \dfrac{1}{4} + \dfrac{1\times 4}{4\times 8} + \dfrac{1\times 4\times 7}{4\times 8\times 12} + \dfrac{1\times 4\times 7\times 10}{4\times 8\times 12\times 16} + \ldots \\ S_2 &=& 1 + \dfrac{2}{6} + \dfrac{2\times 5}{6\times 12} + \dfrac{2\times 5\times 8}{6\times 12\times 18} + \dfrac{2\times 5\times 8\times 11}{6\times 12\times 18\times 24} + \ldots \\ \end{aligned}

Let us define two infinite series, S 1 S_1 and S 2 S_2 as above.

What is the relationship between S 1 S_1 and S 2 S_2 ?

S 1 < S 2 \large S_1< S_2 S 1 > S 2 \large S_1> S_2 S 1 = S 2 \large S_1 = S_2 They both diverge

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1 solution

Rushikesh Joshi
May 4, 2015

See expansion of (1+x)^-n for S1, put x=3/4 and n=1/3. for S2, put x=1/2 and n=2/3. you will get S1=S2=4^(1/3).

A little small for a level five question.

Vishwak Srinivasan - 6 years, 1 month ago

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