Question 8.3.2: The p-Series Determine for which values of p the series ∑^∞k...
The p-Series
Determine for which values of p the series \sum_{k=1}^{\infty} \frac{1}{k^R} (a p-series) converges.
First, notice that for p = 1, this is the harmonic series, which diverges. For
p > 1, define f(x)=\frac{1}{x^p}=x^{-p}. Notice that for x ≥ 1, f is continuous and positive. Further,
so that f is decreasing. This says that the Integral Test applies. We now consider
\int_1^{\infty} x^{-p} d x=\lim _{R \rightarrow \infty} \int_1^R x^{-p} d x=\left.\lim _{R \rightarrow \infty} \frac{x^{-p+1}}{-p+1}\right|_1 ^R=\lim _{R \rightarrow \infty}\left(\frac{R^{-p+1}}{-p+1}-\frac{1}{-p+1}\right)=\frac{-1}{-p+1} . \quad \begin{array}{l}\text { Since } p>1 \text { implies } \\\text { that }-p+1<0\end{array}.
In this case, the improper integral converges and so too, must the series. We leave it as an exercise to show that the series diverges when p < 1.
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