Prove inequality

Question

1/3 < 1/13+1/15+1/17+...+1/61+1/63 < 4/3

Answer 1

Let S = 1/13 + 1/ 15 + 1/17 + ... + 1/61 + 1/63.

Denominators create arithmetic sequence with a₁ = 13, d = 2 and a_n = 63. Then 63 = 13 + 2(n- 1), n = 26.

Then a_n = 2n + 11. If we combine eqwuidistant from the ends we get S = ∑₁¹³(1/(2n+11)+1/(65 - 2n)) = ∑

∑₁¹³(76/(11+2n)(65-2n)) = ∑₁¹³76/(- 4n² + 108n + 715).

Consider function f(x) = 76/(- 4x² + 108x + 715), derivative f'(x) = -76(- 8x + 108)/(-4x²+108x +715)² < 0,

-8x + 108 > 0, x < 13.5. We now prove that 76/(-4n² + 108n + 715) decriasing for n = 1, 2, 3, ..., 13.

This means that 13·76/(2·13 + 11)(65-2·13) < S < 13·76/((2·1 + 11)(65 - 2·1)) or

13·76/(37·39) < S < 13·76/(13·63) or 1/3 < 76/111 < S < 76/63 < 4/3.