Difference between revisions of "The Divergence and Integral Tests"

Latest revision as of 13:15, 29 October 2021

Theorem (p-series)

The p-series $\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}$ converges for $p>1$ and diverges for $0<p\leq 1$ . [Note: We have technically only defined this series for p rational. However, the theorem is still valid when p is irrational, for the same reasons]

Proof

First we'll consider the special cases $p=1,p=2$ , and then obtain the general result from these.

If p = 1, let $x_{n}$ be the greatest power of 2 less than ${\frac {1}{n}}$ . That is, $(x_{n})=({\frac {1}{2}},{\frac {1}{4}},{\frac {1}{4}},{\frac {1}{8}},{\frac {1}{8}},{\frac {1}{8}},{\frac {1}{8}},\dots )$ .

Grouping like terms together, $\sum _{i=1}^{\infty }x_{n}$ $=\sum _{i=0}^{\infty }(\sum _{j=2^{i}}^{2^{i+1}-1}x_{j})$ $=\sum _{i=0}^{\infty }(\sum _{j=2^{i}}^{2^{i+1}-1}{\frac {1}{2^{i+1}}})$ $=\sum _{i=1}^{\infty }{\frac {1}{2}}$ , which diverges.

By definition, $x_{n}<{\frac {1}{n}}$ , so by the comparison test $\sum _{n=1}^{\infty }{\frac {1}{n}}$ diverges.

If p = 2, let $x_{n}={\frac {1}{n(n-1)}}={\frac {1}{n-1}}-{\frac {1}{n}}$ if n>1 and 1 if n=1.

Using the theorem on telescoping series, $\sum _{n=1}^{\infty }x_{n}$ $=1+\sum _{n=2}^{\infty }({\frac {1}{n-1}}-{\frac {1}{n}})$ $=1+1-\lim _{n\rightarrow \infty }(1/n)=2$

By definition, $x_{n}>{\frac {1}{n\cdot n}}={\frac {1}{n^{2}}}$ , so by the comparison test $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ converges as well.

If 0 < p < 1, then ${\frac {1}{n^{p}}}>{\frac {1}{n}}$ . By the comparison test $\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}$ diverges.

Limit Test

The limit test essentially tells us whether or not the series is a candidate for being convergent. It is as follows:

Limit Test for Convergence

If a series $S=\sum _{n}^{\infty }{s_{n}}$ and if $\lim _{n\rightarrow \infty }{s_{n}}\neq 0$ the series must be divergent. If the limit is zero, the test is inconclusive and further analysis is needed.

This follows because if the summand does not approach zero, when $n$ becomes very large, $s_{n}$ will be close to the non-zero $L$ and the series will start behaving like an arithmetic series; remember that arithmetic series never converge.

However, one should not misuse this test. This is a test for divergence and not convergence. A series fails this test if the limit of the summand is zero, not if it is some non-zero $L$ . If the limit is zero, you will need to do other tests to conclude that the series is divergent or convergent.

Example 1

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n+1}{n}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n+1}{n}}=1$

the limit is not zero and so the series is divergent by the limit test.

Example 2

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}={\frac {1}{3}}$

the limit is not zero and so the series is divergent by the limit test.

Example 3

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{n^{3}}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{n^{3}}}=0$

the limit is zero and so the test is inconclusive. Further analysis is needed to determine whether or not the series converges or diverges.

Integral Test

The next test for convergence for infinite series is the integral test. The integral test utilizes the fact that an integral is essentially an Riemann Sum—which is itself an infinite sum—over an infinite interval which is useful because integration is relatively straight forward and familiar.

The test is as follows:

Integral Test for Convergence and Divergence

If a series $S=\sum _{n=j}^{\infty }{s_{n}}$ , and if the discrete function $s(n)$ is continuous and decreasing on the interval $[j,\infty )$ then if

$\int _{j}^{\infty }{s(n)dn}$ is convergent, so is $S$

$\int _{j}^{\infty }{s(n)dn}$ is divergent, so is $S$

Let us look at why this this test works. As an example, we'll use the harmonic series $\sum _{n=1}^{\infty }{\frac {1}{n}}$ . The harmonic series is well known series that actually happens to be divergent. This series actually passes the limit test, however it does not pass the integral test. If we want to approximate the integral, we can use rectangles like in a Riemann Sum:

Note that this right-handed method will always under approximate the integral (assuming that the function is decreasing on our selected interval). This implies that if the right-handed sum is equal to the actual infinite series, then the integral itself must be greater than the sum. This can help show convergence, because if the integral from the starting point to infinity is convergent, then by the comparison test the original function on that interval must also be convergent. And so we can see that the integral test is actually a "special case" of the comparison test.

But what about divergence? This case is also satisfied—if we use a left-handed approximation instead of a right handed one, we see that we again attain the original series, however there is an important difference:

The key difference, in this case, is that the integral becomes an under approximation for the series, and we can use the new "series" of the integral to show divergence with the comparison test.

This test is useful but is unfortunately only useful on functions that can be integrated and are decreasing in size. The latter may seem like a trivial and unnecessary addition, however consider how this test works; it relies on the fact that the integral of a function decreasing on an interval will always yield an under/over approximation of a series; if the function is not decreasing everywhere on the interval, the integral will not necessarily yield an under/over approximation every time.