Question 8.3.1: Using the Integral Test Investigate the convergence or diver...
Using the Integral Test
Investigate the convergence or divergence of the series \sum_{k=0}^{\infty} \frac{1}{k^2+1}.
The graph of the first 20 partial sums shown in Figure 8.24 suggests that the series converges to some value around 2. In the accompanying table, we show some selected partial sums. Based on this, it is difficult to say whether the series is converging
very slowly to a limit around 2.076 or whether the series is instead diverging very slowly. To determine which is the case, we must test the series further. Define f(x)=\frac{1}{x^2+1}. Note that f is continuous and positive everywhere and
f(k)=\frac{1}{k^2+1}=a_k , for all k ≥ 1. Further,
f^{\prime}(x)=(-1)\left(x^2+1\right)^{-2}(2 x)<0,
for x ∈ (0,∞), so that f is decreasing. This says that the Integral Test applies to this series. So, we consider the improper integral
\int_0^{\infty} \frac{1}{x^2+1} d x=\lim _{R \rightarrow \infty} \int_0^R \frac{1}{x^2+1} d x=\left.\lim _{R \rightarrow \infty} \tan ^{-1} x\right|_0 ^R=\lim _{R \rightarrow \infty}\left(\tan ^{-1} R-\tan ^{-1} 0\right)=\frac{\pi}{2}-0=\frac{\pi}{2}.
The Integral Test says that since the improper integral converges, the series must converge, also. Now that we have established that the series is convergent, our earlier calculations give us the estimated sum 2.076. Notice that this is not the same as the value of the corresponding improper integral, which is \frac{\pi}{2} \approx 1.5708.
