CARNEGIE MELLON UNIVERSITY
DEPARTMENT OF MATHEMATICAL SCIENCES
Integral Test Approximation
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 estimate is sometimes not a very good estimate, and may require many terms for a desired approximation. To improve matters we define an approximation which involves both a partial sum and an improper integral.

$$A_n=s_n+ \frac{1}{2} \left( I_{n+1}+I_n \right).$$ (3)

The error in using $A_n$ to approximate the actual sum $s$ is given by $s-A_n$. Now

$$\begin{eqnarray*} \vert s-A_n\vert & = & \vert s-s_n- (1/2)\left( I_{n+1}+I_n \r... ...infty f(x) dx \right) \\ & = & (1/2) \int_n^{n+1} f(x) dx. \end{eqnarray*}$$

We now simplify things by using the fact that $f$ is decreasing, so $f(x) \leq f(n)$ on the interval $[n,n+1]$. We obtain

$$\vert s-A_n\vert \leq \frac{1}{2} a_n.$$ (4)

This can be used to determine $n$ so that $A_n$ is accurate to within the desired amount $\epsilon$. Perhaps a slightly smaller value could work - but this is usually not so important. Try the following exercises where we apply the above.

1. Approximate $\sum_{n=1}^{\infty} \frac{1}{n^4}$ to within $\epsilon=0.001$ using (4) to determine the value $n$ to use in computing $A_n$ in (3).
2. Use the fact that $s=\pi^4/90$ in the above to determine the actual error, $s-A_n$, for the approximation $A_n$ that you computed.
3. Now determine the actual error when using $s_n$, for your value of $n$.
4. How many terms do you need to approximate $s$ to within $\epsilon=0.001$ using only partial sums $s_n$?