Given remainder estimate of 10^-10, what is the fewest number of terms of the series you would need?

Question

Answer 1

Let f(x) = 9/x². x > 0. This is positive and decreasing function Then remainder R_n < ∫_n^∞1/x²dx= lim a → ∞∫

∫_n^a9/x²dx = lim a → ∞ (- 9/x)_n^a = 9/n< 10^-10; n > 9·10¹⁰

S_n + ∫_n+1^∞9/x²dx < ∑₁^∞9/n² < S_n + ∫_n^∞9/x²dx;

S_n + 9/(n+1) < S < Sn + 9/n ; 9/(n + 1) < S - S_n< 9/n; 9/n - 9/(n+1) < 10^-10;

9/[n(n+1)] < 10^-10; 9/n² < 10^-10; n² > 9·10¹⁰; n > 3·10⁵

Answer 2

[9 / (n + 1)²] - [9 / n² ] < 10^-10

Can you solve for n and answer?