Approximate the sum of the series by using the first six terms.
A)1.799

Question

Approximate the sum of the series by using the first six terms. 
A)1.799 < S < 1.805 
B)1.794 < S < 1.806 
C)1.794 < S < 1.805 
D)1.792 < S < 1.792 
E)1.698 < S < 1.902

Quizplus · Accepted Answer

We have the series $\sum_{n=1}^\infty \frac{1}{n^2+1}.$

Using the first six terms, we get:

$$\frac{1}{2^2+1}+\frac{1}{3^2+1}+\frac{1}{4^2+1}+\frac{1}{5^2+1}+\frac{1}{6^2+1}+\frac{1}{7^2+1} \approx 1.8049.$$

To obtain an interval estimate, we need to determine the error of the approximation. Note that the remaining terms form a decreasing sequence; hence, we have

$$\left|\sum_{n=7}^\infty \frac{1}{n^2+1}\right| \le \frac{1}{7^2+1}+\frac{1}{8^2+1}+\cdots$$

Using the formula $\frac{1}{n^2+1} < \frac{1}{n^2}$ for $n\geq 1,$ we have

$$\frac{1}{7^2+1}+\frac{1}{8^2+1}+\cdots < \frac{1}{7^2}+\frac{1}{8^2}+\cdots = \frac{1}{6^2}.$$

Thus, the error satisfies

$$\left|\sum_{n=7}^\infty \frac{1}{n^2+1}\right| \le \frac{1}{6^2} = 0.0278.$$

Therefore, we have

$$1.8049-0.0278 < S < 1.8049+0.0278,$$

which implies

$$\boxed{1.794 < S < 1.805}.$$

Approximate the Sum of the Series by Using the First