p series
p series are infinite series of the form
$$\sum_{n = 1}^{\infty} \, \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \dots + \frac{1}{n^p}$$
where $n$ in the denominator is what changes with the index of the summation, and the power to which that denominator is raised is $p$.
Below we will show that a p-series converges for $p \gt 1$, and diverges otherwise. An important divergent p-series is the harmonic series,
$$\sum_{n = 1}^{\infty} \, \frac{1}{n^p} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$$
Notice that we begin the index of a p-series with $n = 1$ because we don't want to have a zero denominator.
The p-integral
The p-integral is the integral of $f(x) = x^{-p}$, were $p$ is a constant. For $p=1$ we know that the integral is $ln|x| + C$, but in order to examine the convergence or divergence of p-series.
The integral is a simple one (I've flipped the exponent and denominator of the integral from $-p+1$ to $1-p$ for convenience).
$$\lim_{R\to\infty} \int_a^R x^{-p} \, dx = \lim_{R\to\infty} \, \frac{x^{1 - p}}{1 - p} \bigg|_a^R$$
Now of course this is an improper integral, so we have to evaluate it using the limit as our "dummy" variable $R$ approaches $\infty$.
The second term above is finite because a is a constant. The first is trickier. Depending on the value of $p$, it will either converge to a limit or diverge to infinity.
Here's the full limit for the case where $p \gt 1$:
$$\lim_{R\to\infty} \int_a^R \, x^{-p} \, dx = \frac{a^{1 - p}}{1 - p}$$
and for $p \lt 1$:
$$\lim_{R\to\infty} \int_a^R \, x^{-p} \, dx \; \longrightarrow \; \infty$$
We already said that when $p=1$ the integral is $ln|x| + C$, which approaches $\infty$ as $R \rightarrow \infty$. With the p-integral in hand, we can use the integral test to determine which p-series converge.
So a p-series
converges if $p \gt 1$ otherwise, it diverges
Convergence of a p-series
A p-series converges if $p \gt 1$, otherwise it diverges.
The harmonic series
The harmonic series is
$$\sum_{n = 1}^{\infty} \frac{1}{x} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$$
To determine whether the harmonic series converges, we let $f(x) = 1/x$ and integrate the function from 1 to $\infty$. Clearly the series is trapped under $f(x)$ as drawn, so if the function converges to a finite limit, so must the series. On the other hand, if the function diverges, then the series won't converge to a fixed limit either.
This integral is the special case of the p-integral with $p = 1$. It goes like this:
$$ \begin{align} \lim_{R\to\infty} \int_1^R \frac{1}{x} \, dx &= \lim_{R\to\infty} ln|x| \, \bigg|_1^R \\[5pt] &= \lim_{R\to\infty} ln|R| - ln(1) \; \rightarrow \; \infty \end{align}$$
Because the integral that traps the series below it does not reach a finite limit (diverges), the series diverges.
The harmonic series is one that causes a lot of confusion for students new to series. As $n \rightarrow \infty$, the size of the term clearly approaches zero, yet the sum doesn't converge (see the comparison with a convergent p-series below.) The trouble with the harmonic series is that the terms just don't approach zero fast enough. If added enough terms of the harmonic series, you could get to any sum, although for large numbers, you might end up adding a vast number of terms.
Comparison of the harmonic series with a convergent p-series
The denominator of the terms of the harmonic series just doesn't get large "fast enough" for the series to converge. Here is a table of terms of the harmonic series and the convergent p-series with terms of the form $1/n^2$. It shows that
Notice that the sum of the terms of the $1/x$ sum just keeps growing, albeit more slowly as x increases. But the $1/x^2$ sum reaches an asymptote below 1.6. At some point, each digit to the right of the decimal point becomes "locked in" in the sum, and doesn't change any more. That's not true of the $1/x$ sum.
This graph might help illustrate the point:
The difference in the two graphs is the presence of a horizontal asymptote on the $1/x^2$ series – a limit on its growth.
Practice problems
Decide whether the following series are p-series, and if so, whether they converge.
-
$$\sum_{n = 1}^{\infty} \frac{n^2}{n^4}$$
Solution
$$\sum_{n = 1}^{\infty} \, \frac{n^2}{n^4} \; \; \text{ reduces to } \; \; \sum_{n = 1}^{\infty} \, \frac{1}{n^2}$$
a p-series with $(p = 2) \gt 1$, so this is a convergent p-series.
-
$$\sum_{n = 1}^{\infty} \frac{1}{\sqrt{1 + n}}$$
Solution
$$\sum_{n = 1}^{\infty} \, \frac{1}{\sqrt{1 + n}}$$
Let u = n + 1, and substitute:
$$\sum_{n = 2}^{\infty} \, \frac{1}{\sqrt{u}} = \sum_{n = 2}^{\infty} \, \frac{1}{u^{1/2}}$$
Notice that we changed the lower limit of integration to match the variable change $x \rightarrow u$. This doesn't affect the outcome of our test because we only care about what happens as $x \rightarrow \infty$
This is a divergent p-series with $p \lt 1$.
-
$$\sum_{n = 1}^{\infty} \frac{\sec^2(n) - \tan^2(n)}{n}$$
Solution
$$\sum_{n = 1}^{\infty} \, \frac{\sec^2(n) - \tan^2(n)}{n}$$
This series is just the harmonic series disguised. Notice that $\sec^2(n) - \tan^2(n) = 1$, is one of the Pythagorean identities. So this series is really the divergent p-series:
$$\sum_{n = 1}^{\infty} \, \frac{1}{n}$$
-
$$\sum_{n = 1}^{\infty} \frac{n}{n^3 + 5}$$
Solution
$$\sum_{n = 1}^{\infty} \, \frac{n}{n^3 + 5}$$
In the limit of large n, the 5 doesn't mean much, making this series approximately equal to the convergent p-series $\sum \frac{1}{n^2}.$
For this series, we'll jump to the comparison test, which you may or may not have seen. It's pretty straightforward, though. We know that the series with terms 1/n2 converges, and it's reasonable to think that the terms of the current series are, term-by-term, smaller than those. If that's true, our series must converge.
$$\frac{n}{n^3 + 5} \lt \frac{n}{n^3} = \frac{1}{n^2}$$
Cross multiplying the inequality gives:
$$n^3 \lt n^3 + 5$$
This is true for all n, so
this series converges.
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