Estimating the Value of a Series — Question 9

Question 9 – Solution

Step 1: Use cancellation.

Since $1/n^2$ decreases to $0$, the alternating estimate gives $$|S-S_N|\le\frac1{(N+1)^2}.$$

Step 2: Ignore signs and bound the absolute tail.

$$\sum_{n=N+1}^{\infty}\frac1{n^2}\le\int_N^{\infty}\frac{dx}{x^2}=\frac1N.$$ This also bounds the signed error, but it discards cancellation and is therefore much weaker.

Step 3: Compare numerically.

At $N=100$, $$\frac1{101^2}=\frac1{10201}\approx9.80\times10^{-5},\qquad \frac1{100}=10^{-2}.$$ The alternating bound is over $100$ times smaller. In general, it is order $N^{-2}$ rather than $N^{-1}$.