Estimating the Sum of a Series

Suppose $\sum a_n$ is a convergent sequence of positive integers. Let

$$R_m = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{m} a_n = \sum_{n=m+1}^{\infty} a_m$$

which is the error if you approximate $\sum a_n$ using the first $n$ terms. From Theorem 6.3.2 we get the following.

Proposition 6.3.8 (Remainder Bound)   Suppose $f$ is a continuous, positive, decreasing function on $[m,\infty)$ and $\sum a_n$ is convergent. Then

$$\int_{m+1}^{\infty} f(x) dx \leq R_m \leq \int_{m}^{\infty} f(x) dx.$$

Proof. In Theorem 6.3.2 set $k=m+1$. That gives

$$\int_{m+1}^{\infty} f(x) dx \leq \sum_{n=m+1}^{\infty} a_n \leq a_{m+1} + \int_{m+1}^{\infty} f(x) dx.$$

But

$$a_{m+1} + \int_{m+1}^{\infty} f(x) dx \leq \int_{m}^{\infty} f(x) dx$$

since $f$ is decreasing and $f(m+1)=a_{m+1}$. $\qedsymbol$

Example 6.3.9   Estimate $\zeta(3) = \sum_{n=1}^{\infty} \frac{1}{n^3}$ using the first $10$ terms of the series. We have

$$\sum_{n=1}^{10} = \frac{19164113947}{16003008000} = 1.197531985674193\ldots$$

The proposition above with $m=10$ tells us that

$$0.00413223140495867\ldots = \int_{11}^{\infty} \frac{1}{x^3}dx \l... ...{10}^{\infty} \frac{1}{x^3}dx = \frac{1}{2\cdot 10^2} = \frac{1}{200} = 0.005.$$

In fact,

$$\zeta(3) = 1.202056903159594285399738161511449990\ldots$$

and we hvae

$$\zeta(3) - \sum_{n=1}^{10} = 0.0045249174854010\ldots,$$

so the integral error bound was really good in this case.

Example 6.3.10   Determine if $\sum_{n=1}^{\infty} \frac{2006}{117n^2 + 41n + 3}$ convergers or diverges. Answer: It converges, since

$$\frac{2006}{117n^2 + 41n + 3} \leq \frac{2006}{117n^2} = \frac{2006}{117} \cdot \frac{1}{n^2},$$

and $\sum \frac{1}{n^2}$ converges.