CARNEGIE MELLON UNIVERSITY
DEPARTMENT OF MATHEMATICAL SCIENCES
Integral Test Approximation Methods
Dr. Timothy Flaherty

Let $\sum_{n=1}^{\infty} a_n$ be a series which converges by the integral test, with $f(x)$ satisfying $f(n)=a_n$ for all $n$. We wish to estimate the actual sum $s$ of the series. First approximate $s$ with a partial sum $s_n=\sum_{k=1}^n a_k$, with error given by the remainder

$$R_n=s-s_n.$$ (1)

As shown in the text,

$$\int_{n+1}^{\infty} f(x) dx \leq R_n \leq \int_n^{\infty} f(x) dx.$$ (2)

Let $I_n=\int_n^{\infty} f(x) dx$, and $I_{n+1}=\int_{n+1}^{\infty} f(x) \ dx$, so $I_{n+1} \leq R_n \leq I_n$. This produces Method 1. We let $A_n$ be the approximation, and set $E_n=s-A_n$ the error.

Method 1 $s \approx A_n$, where $A_n=s_n$, and $\vert E_n\vert\leq I_n$.

Handout 1 on this topic establishes:

Method 2 $s \approx A_n$, where $A_n=s_n+\frac{1}{2}(I_{n}+I_{n+1})$, with $\vert E_n\vert\leq (1/2)a_n$.

The third method is derived by estimating the integral of $f(x)$ on $[n,n+2]$. An upper bound is obtained by finding the area beneath the segment joining $(n,a_n)$ and $(n+2,a_{n+2})$. A lower bound is obtained from the trapezoid with top segment given by the tangent to $f(x)$ at $(n+1, a_{n+1})$. So the integral is bounded between $2a_{n+1}$ and $a_{n}+a_{n+2}$. Doing this on all intervals of the form $[n+k,n+k+2]$, adding and regrouping produces

Method 3 $s \approx A_n$, where $A_n=s_n-\frac{1}{4}(a_n-a_{n+1})+I_n-(1/2)\int_n^{n+1} f(x) dx$, with $\vert E_n\vert\leq (1/4)(a_n-a_{n+1}) \leq -(1/4)f^{\prime}(n)$.

The forth method is obtained by estimating the integral of $f(x)$ on $[n,n+1]$. An upper bound is obtained by finding the area beneath the segment joining $(n,a_n)$ and $(n+1, a_{n+1})$. A lower bound is obtained from two trapezoids. The first is above $[n,n+1/2]$ with top segment the tangent at $(n,a_n)$. The second trapezoid is above $[n+1/2,n+1]$ with top segment the tangent at $(n+1,f^{\prime}(n+1))$. These estimates produce:

Method 4 $s \approx A_n$, where $A_n=s_n+I_n-(1/2)a_n-(1/16)f^{\prime}(n)$, with $\vert E_n\vert\leq -(1/16)f^{\prime}(n)$.

As a comparison, for the p-series with $p=1.01$, and with error $\epsilon \leq 0.001$, we require $10^{500}$, $470$, $16$, and $8$ terms to approximate the sum using methods 1,2,3, and 4, respectively. To within $10^{-6}$ the actual sum is 10.577943.