inf
--- 1
\ ----
/___ ln(n)
n=2
Diverges, even though the lim
n->inf1/ln(n) = 0...
nth-Term Test for Divergence
If a sequence {an} does not converge to 0, then the series
Sigma(an) diverges. This does not imply that if the sequence {an} converges to 0 that the series Sigma(an) converges; the test is merely inconclusive. In other words...
inf
--- n
\ ----
/___ 2
n=1
Diverges because lim
n->inf n/2 = infinity, which does not exist.
Geometric Series
Geometric series are rather simple when it comes to determining their convergence or divergence. First, it would be best to define geometric series.
inf
---
\ ar^n = a + ar + ar^2 + ar^3 +... ar^n +..., r != 0
/___
n=0
The r term is referred to as the
ratio. If 0 < |r| < 1, then the geometric series converges. To determine the
series, if it converges, simply follow the formula S = a/(1-r).
Integral Test
This is another rather simple test that works wonderfully for all easily integrable problems. If f is a positive, continuous, decreasing function for n >= 1, and if f(n) = an, then both
|\ infinity
| f(x) dx and
\| 1
inf
---
\ an
/___
n=0
either converge or diverge.
P-Series
P-series are are special series defined as follows:
inf
---
\ 1/n^p = 1/1^p + 1/2^p +... 1/n^p
/___
n=1
To figure out the divergence or convergence of these series, just look at the
p. If 0 < p <= 1, then the series diverges. If p > 1, the
series converges.
Comparison Tests
If an <= cn for all n, and
inf
---
\ cn converges, then
/___
n=1
inf
---
\ an converges
/___
n=1
And if c
n <= a
n,
inf
---
\ cn diverges, then
/___
n=1
inf
---
\ an diverges
/___
n=1
In a similar light, suppose a
n > 0 and d
n > 0. lim
n-> d
n/a
n = L, where L is finite and positive, then the series
inf
---
\ dn and
/___
n=1
inf
---
\ an
/___
n=1
both either converge or diverge.
There are several other tests for divergence and convergence, but the preceeding tests are a solid foundation for dealing with infinite series.