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A series usually defined as the sum of the terms in an infinite sequence. A series is considered convergent if the limit of the series approaches a specific value and divergent if it approaches positive or negative infinity.

Divergence and convergence tests

Geometric series

If a series is geometric, or in the form

\sum_{n=0}^\infty ar^n

it will be convergent if |r|<1 . The sum will be equal to

\frac{a}{1-r}

Comparison test

If a series converges, and every term in a second series is smaller than all corresponding terms in the first series, the second series must also converge. The opposite is also true (if the first series diverges and all the terms in the second are larger, the second series diverges).

Limit comparison test

If a_n,b_n>0 , \lim_{n\to\infty}\frac{a_n}{b_n} is finite and not equal to 0, and if and only if \sum_{n=1}^\infty b_n converges, \sum_{n=1}^\infty a_n will converge.

Limit test

If the limit of a sequence being summed does not equal 0, the series diverges.

Root test

If the limit L=\lim_{n\to\infty}\sqrt[n]{|a_n|}<1 , the series \sum_{n=1}^\infty a_n is convergent. If it is greater than one, it will diverge, and if it is equal to one, it can diverge or converge.

Integral test

If f(n)=a_n and -\infty<\int\limits_a^\infty f(x)dx<\infty , \sum_{n=a}^\infty a_n must converge. It is important to note that

\int\limits_a^\infty f(x)dx\ne\sum_{n=a}^\infty a_n

Ratio test

Given the series \sum_{n=k}^\infty a_n , and \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=r

  1. If r<1 the series converges.
  2. If r>1 the series diverges.
  3. If r=1 the series could be either divergent or convergent.

This test is also known as d'Alembert's ratio test or as the Cauchy ratio test.

Alternating series test

An alternating series, or one in the form \sum_{n=1}^\infty(-1)^n a_n or \sum_{n=1}^\infty(-1)^{n-1}a_n will converge if

  1. a_n\ge a_{n+1} for all n
  2. \lim_{n\to\infty}a_n=0

Interval of convergence

If the terms of the sum have a variable in them, the interval over which the sum converges is called the interval of convergence. This interval can be found by taking the ratio test \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=r . The interval of convergence is the interval of the variable in which |r|<1 .

For example,

\sum_{n=k}^\infty\frac{1}{nx^n}
r =\lim_{n\to\infty}\left|\frac{\dfrac{1}{(n+1)x^{n+1}}}{\dfrac{1}{nx^n}}\right|=\lim_{n\to\infty}\left|\frac{nx^n}{(n+1)x^{n+1}}\right|=\lim_{n\to\infty}\left|\frac{n}{(n+1)x}\right|
=\left|\frac{1}{x}\right|\lim_{n\to\infty}\left|\frac{n}{(n+1)}\right|=\left|\frac{1}{x}\right|\cdot1=\left|\frac{1}{x}\right|
|r|<1 when |x|>1 , so the interval of convergence is x<-1,x>1 .

This test does not tell us anything about the endpoints, so they must be tested separately.

Absolute versus conditional convergence

A series is said to converge conditionally if the sum converges but the sum of the absolute value of the terms does not, and a series converges absolutely if the absolute value of the terms converges as well. Since conditionally convergent series are most often seen in the form of alternating series, the alternating series test is typically the most useful in finding the type of convergence.

Applications

One of the most important applications of series are Taylor approximations, which can be used to approximate transcendental functions and transcendental numbers. Another important series is the Fourier series, which is used to approximate periodic functions.

Examples

Does the series below converge?

\sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots

We can use the integral test here. Since \int\limits_a^\infty f(x)dx=\int\limits_a^\infty\frac{dx}{n}=\infty , this series diverges. This series is known as the harmonic series.


Does the series below converge?

\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots

Now we can use the alternating series test. Since a_n\ge a_{n+1} for all n , and \lim_{n\to\infty}\frac{1}{n}=0 , this series converges, although not absolutely, since

\sum_{n=1}^\infty\left|\frac{(-1)^{n-1}}{n}\right|=\sum_{n=1}^\infty\frac{1}{n}=\infty

See also

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