P Series

Decide if the series

$\sum _{n=1}^{\infty }\frac{\sqrt[4]{n}}{n^5}$

is a p-series or not.

Answer:

It doesn't look like a p-series at first glance, but let's do some algebra just to be sure. Here

$\frac{\sqrt[4]{n}}{n^5}=\frac{n^{\frac{1}{4}}}{n^5}=\frac{1}{n^5·n^{-\frac{1}{4}}}=\frac{1}{n^{5-\frac{1}{4}}}=\frac{1}{n^{\frac{19}{4}}}={\left(\frac{1}{n}\right)}^{\frac{19}{4}}$

so in fact this is a p-series.

Is the series

$\sum _{n=1}^{\infty }\frac{1}{3^n}$

a p-series?

Answer:

No, because in a p-series it needs to be $\frac{1}{n}$ raised to a constant power, not a constant raised to the$n$power. In fact this series also has a special name, it is called a Geometric series. For more information on this type of series see Geometric Series.

Does the series

$\sum _{n=1}^{\infty }\frac{1}{n·\sqrt[3]{n}}$

converge or diverge?

Answer:

While this may not look like a p-series, remember that

$\frac{1}{n·\sqrt[3]{n}}=\frac{1}{n·n^{\frac{1}{3}}}=\frac{1}{n^{\frac{4}{3}}}$,

so this is actually a p-series. To use the Integral Test, take the function

$f\left(x\right)=\frac{1}{x^{\frac{4}{3}}}$ .

This function is decreasing, continuous, and positive for which means the conditions are met to apply the Integral Test. Then integrating,

$$\begin{gathered} {\int }_1^{\infty }f\left(x\right)dx={\int }_1^{\infty }\frac{1}{x^{\frac{4}{3}}}dx \\ ={\int }_1^{\infty }{x}^{-\frac{4}{3}}dx \\ =\underset{k\to \infty }{\lim }{\int }_1^k{x}^{-\frac{4}{3}}dx \\ =\underset{k\to \infty }{\lim }\left[{\overline{)-3x^{-\frac{1}{3}}}}_1^k\right] \\ =\underset{k\to \infty }{\lim }\left[-3k^{-\frac{1}{3}}-\left(-3\right)\right] \\ =3. \end{gathered}$$

Since the integral converges, by the Integral Test the series converges too.

Does the series

$\sum _{n=1}^{\infty }\frac{1}{n^2}$

converge or diverge?

Answer:

In this example$p=2$. Since $p>\hspace{0.17em}1$, this series converges.

Explain why the series $\sum _{n=1}^{\infty }{\left(\frac{1}{n}\right)}^{-2}$ diverges by writing out some of the partial sums and showing what happens.

Answer:

Writing out the partial sums,

$\begin{array}{ccc}{s}_1& =& {\left(\frac{1}{1}\right)}^{-2}=1\\ s_2& =& s_1+{\left(\frac{1}{2}\right)}^{-2}=1+4=5\\ s_3& =& s_{2\hspace{0.17em}+}{\left(\frac{1}{3}\right)}^{-2}=5+\hspace{0.17em}9=14,\end{array}$

and as you can see they just keep getting bigger. That means the series diverges.

Decide if the series

$\sum _{n=1}^{\infty }\frac{2^n}{n^2{3}^n}$

converges or diverges.

Answer:

First let's rewrite the series in question as

$\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^n\frac{1}{n^2}$ .

You already know that the series

$\sum _{n=1}^{\infty }\frac{1}{n^2}$

is a p-series with$p=2$, and it converges. It also has all positive terms. So if you can squash the terms of the series you are looking at (which also has all positive terms) under the terms of the p-series, you can use the Comparison Test to say that the new series converges.

Looking at

${\left(\frac{2}{3}\right)}^n$,

because $n\ge 1$ you can say that

${\left(\frac{2}{3}\right)}^n<1$.

So

${\left(\frac{2}{3}\right)}^n{\left(\frac{1}{n}\right)}^2<\hspace{0.17em}1·{\left(\frac{1}{n}\right)}^2={\left(\frac{1}{n}\right)}^2$.

That means that by the Comparison Test using the p-series with , the series

$\sum _{n=1}^{\infty }\frac{2^n}{n^2{3}^n}$

converges.