Integral test and p-series

One good way to get an upper bound for the partial sums of a series is with an integral:

Theorem 6.1 (Integral Test)   If a_n=f(n) for every $n\in \ensuremath{\mathbb{N} }$ and if f is a positive, decreasing function, then $\sum a_n$ converges if and only if $\int_0^\infty f(x) \ dx$ converges.

PROOF:

By constructing step functions based on the series we can see that

$$\sum_{n=1}^k a_n < \int_0^k f(x)\ dx < \sum_{n=0}^{k-1}a_k$$

Now if $\int_0^\infty f(x) \ dx$ converges then we get

$$\sum_{n=0}^k a_n < a_0+\int_0^\infty f(x) \ dx$$

giving a bound on the partial sums. Thus $\sum a_n$ converges.

If, on the other hand, $\int_0^\infty f(x) \ dx$ diverges then the numbers $\int_0^k f(x)\ dx$ get arbitrarily large, and thus so do the partial sums. This tells us that the series also diverges.

Corollary 6.2 (p-series)   The series

$$\sum_{n=1}^\infty \frac1{n^p}$$

converges if and only if p>1.

PROOF:

The integral

$$\int_1^\infty \frac1{x^p} \ dx=\lim_{U\to \infty}\frac{U^{-p+1}}{-p+1}-\frac1{1-p}.$$

For p>1 this converges to $\frac1{p-1}$. For p<1 it diverges. If p=1 then the integral gives a natural $\ln(U)$ which diverges as $U\to \infty$.

Examples:

1.

$\sum\frac1{n^2}$ converges

2.

$\sum\frac1{\sqrt{n}}$ diverges

3.

$\sum\frac1n$ diverges. This is called the harmonic series.

Usually we use the limit comparison theorem with p-series when checking endpoints of intervals of convergence:

Theorem 6.3 (Limit Comparison Theorem)   If 0<a_n and 0<b_n and

$$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$

then if $0<L<\infty$ then either both of $\sum a_n$ and $\sum b_n$ converge or both diverge. If L=0 and $\sum b_n$ converges, then so does $\sum a_n$.

PROOF:

Suppose that $L<\infty$ and $\sum b_n$ converges. Then for large enough n we know that

$$\frac{a_n}{b{n}}<L+1$$

thus a_n<(L+1)b_n. Now if $\sum b_n$ converges then so does $\sum (L+1)b_n$ so the comparison theorem will tell us that $\sum a_n$ converges.

If L>0 we can invert to see that

$$\lim_{n\to \infty} \frac{b_m}{a_n}\to \frac{1}{L}<\infty$$

so if $\sum a_n$ converges, so does $\sum b_n$.

Examples:

1.

$${\sum \frac{n-1}{n^3+n^2-n}}$$ converges by limit comparison with $${\sum\frac1{n^2}}$$. Here

$$\lim_{n\to \infty}\frac{\frac{n-1}{n^3+n^2-n}}{\frac1{n^2}}=\lim_{n\to\infty}\frac{n^3-n^2}{n^3+n^2-n}=1$$

2.

$${\sum \frac{n-1}{n^2-n}}$$ diverges by limit comparison with $${\sum\frac1{n}}$$. Here

$$\lim_{n\to \infty}\frac{\frac{n-1}{n^2-n}}{\frac1{n}}=\lim_{n\to\infty}\frac{n^2-n}{n^2-n}=1$$

Exercises: Here are some for you to try:

1.

$${\sum\frac{n}{\sqrt{n^3-n}}}$$

2.

$${\sum\frac{n^2}{n^3-n}}$$

3.

$${\sum\frac{\sqrt{n}}{n^3-n}}$$

4.

$${\sum\frac{\sqrt{n}}{\sqrt{n^3-n}}}$$

5.

$${\sum\frac{3n}{n^3-n^2}}$$