Misplaced Pages

Dirichlet's test

Article snapshot taken from Wikipedia with creative commons attribution-sharealike license. Give it a read and then ask your questions in the chat. We can research this topic together.
Test for series convergence
Part of a series of articles about
Calculus
a b f ( t ) d t = f ( b ) f ( a ) {\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}
Differential
Definitions
Concepts
Rules and identities
Integral
Definitions
Integration by
Series
Convergence tests
Vector
Theorems
Multivariable
Formalisms
Definitions
Advanced
Specialized
Miscellanea

In mathematics, Dirichlet's test is a method of testing for the convergence of a series that is especially useful for proving conditional convergence. It is named after its author Peter Gustav Lejeune Dirichlet, and was published posthumously in the Journal de Mathématiques Pures et Appliquées in 1862.

Statement

The test states that if ( a n ) {\displaystyle (a_{n})} is a monotonic sequence of real numbers with lim n a n = 0 {\textstyle \lim _{n\to \infty }a_{n}=0} and ( b n ) {\displaystyle (b_{n})} is a sequence of real numbers or complex numbers with bounded partial sums, then the series

n = 1 a n b n {\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}

converges.

Proof

Let S n = k = 1 n a k b k {\textstyle S_{n}=\sum _{k=1}^{n}a_{k}b_{k}} and B n = k = 1 n b k {\textstyle B_{n}=\sum _{k=1}^{n}b_{k}} .

From summation by parts, we have that S n = a n B n + k = 1 n 1 B k ( a k a k + 1 ) {\textstyle S_{n}=a_{n}B_{n}+\sum _{k=1}^{n-1}B_{k}(a_{k}-a_{k+1})} . Since the magnitudes of the partial sums B n {\displaystyle B_{n}} are bounded by some M and a n 0 {\displaystyle a_{n}\to 0} as n {\displaystyle n\to \infty } , the first of these terms approaches zero: | a n B n | | a n M | 0 {\displaystyle |a_{n}B_{n}|\leq |a_{n}M|\to 0} as n {\displaystyle n\to \infty } .

Furthermore, for each k, | B k ( a k a k + 1 ) | M | a k a k + 1 | {\displaystyle |B_{k}(a_{k}-a_{k+1})|\leq M|a_{k}-a_{k+1}|} .

Since ( a n ) {\displaystyle (a_{n})} is monotone, it is either decreasing or increasing:

  • If ( a n ) {\displaystyle (a_{n})} is decreasing, k = 1 n M | a k a k + 1 | = k = 1 n M ( a k a k + 1 ) = M k = 1 n ( a k a k + 1 ) , {\displaystyle \sum _{k=1}^{n}M|a_{k}-a_{k+1}|=\sum _{k=1}^{n}M(a_{k}-a_{k+1})=M\sum _{k=1}^{n}(a_{k}-a_{k+1}),} which is a telescoping sum that equals M ( a 1 a n + 1 ) {\displaystyle M(a_{1}-a_{n+1})} and therefore approaches M a 1 {\displaystyle Ma_{1}} as n {\displaystyle n\to \infty } . Thus, k = 1 M ( a k a k + 1 ) {\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})} converges.
  • If ( a n ) {\displaystyle (a_{n})} is increasing, k = 1 n M | a k a k + 1 | = k = 1 n M ( a k a k + 1 ) = M k = 1 n ( a k a k + 1 ) , {\displaystyle \sum _{k=1}^{n}M|a_{k}-a_{k+1}|=-\sum _{k=1}^{n}M(a_{k}-a_{k+1})=-M\sum _{k=1}^{n}(a_{k}-a_{k+1}),} which is again a telescoping sum that equals M ( a 1 a n + 1 ) {\displaystyle -M(a_{1}-a_{n+1})} and therefore approaches M a 1 {\displaystyle -Ma_{1}} as n {\displaystyle n\to \infty } . Thus, again, k = 1 M ( a k a k + 1 ) {\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})} converges.

So, the series k = 1 B k ( a k a k + 1 ) {\textstyle \sum _{k=1}^{\infty }B_{k}(a_{k}-a_{k+1})} converges by the direct comparison test to k = 1 M ( a k a k + 1 ) {\textstyle \sum _{k=1}^{\infty }M(a_{k}-a_{k+1})} . Hence S n {\displaystyle S_{n}} converges.

Applications

A particular case of Dirichlet's test is the more commonly used alternating series test for the case b n = ( 1 ) n | n = 1 N b n | 1. {\displaystyle b_{n}=(-1)^{n}\Longrightarrow \left|\sum _{n=1}^{N}b_{n}\right|\leq 1.}

Another corollary is that n = 1 a n sin n {\textstyle \sum _{n=1}^{\infty }a_{n}\sin n} converges whenever ( a n ) {\displaystyle (a_{n})} is a decreasing sequence that tends to zero. To see that n = 1 N sin n {\displaystyle \sum _{n=1}^{N}\sin n} is bounded, we can use the summation formula n = 1 N sin n = n = 1 N e i n e i n 2 i = n = 1 N ( e i ) n n = 1 N ( e i ) n 2 i = sin 1 + sin N sin ( N + 1 ) 2 2 cos 1 . {\displaystyle \sum _{n=1}^{N}\sin n=\sum _{n=1}^{N}{\frac {e^{in}-e^{-in}}{2i}}={\frac {\sum _{n=1}^{N}(e^{i})^{n}-\sum _{n=1}^{N}(e^{-i})^{n}}{2i}}={\frac {\sin 1+\sin N-\sin(N+1)}{2-2\cos 1}}.}

Improper integrals

An analogous statement for convergence of improper integrals is proven using integration by parts. If the integral of a function f is uniformly bounded over all intervals, and g is a non-negative monotonically decreasing function, then the integral of fg is a convergent improper integral.

Notes

  1. Démonstration d’un théorème d’Abel. Journal de mathématiques pures et appliquées 2nd series, tome 7 (1862), pp. 253–255 Archived 2011-07-21 at the Wayback Machine. See also .
  2. ^ Apostol 1967, pp. 407–409
  3. Spivak 2008, p. 495
  4. ^ Rudin 1976, p. 70
  5. Rudin 1976, p. 71
  6. "Where does the sum of $\sin(n)$ formula come from?".

References

External links

Calculus
Precalculus
Limits
Differential calculus
Integral calculus
Vector calculus
Multivariable calculus
Sequences and series
Special functions
and numbers
History of calculus
Lists
Integrals
Miscellaneous topics
Category: