Test for series convergence
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In mathematics, Dirichlet's test is a method of testing for the convergence of a series. It is named after its author Peter Gustav Lejeune Dirichlet, and was published posthumously in the Journal de Mathématiques Pures et Appliquées in 1862.[1]
Statement
The test states that if
is a sequence of real numbers and
a sequence of complex numbers satisfying
is monotonic ![{\displaystyle \lim _{n\to \infty }a_{n}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42e41a1150b9009115ca85dfbcda86b4586bcd12)
for every positive integer N
where M is some constant, then the series
![{\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1da4118b971026424454fb80f2458ef9b4cac33)
converges.
Proof
Let
and
.
From summation by parts, we have that
. Since
is bounded by M and
, the first of these terms approaches zero,
as
.
We have, for each k,
.
Since
is monotone, it is either decreasing or increasing:
- If
is decreasing, ![{\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}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71607b6d437ebc9c42cb1f5e0d555319e94b074f)
which is a telescoping sum that equals
and therefore approaches
as
. Thus,
converges. - If
is increasing, ![{\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}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/068b2c4fc2372410a8404d9a32d6b14abe6d6924)
which is again a telescoping sum that equals
and therefore approaches
as
. Thus, again,
converges.
So, the series
converges, by the absolute convergence test. Hence
converges.
Applications
A particular case of Dirichlet's test is the more commonly used alternating series test for the case
![{\displaystyle b_{n}=(-1)^{n}\Longrightarrow \left|\sum _{n=1}^{N}b_{n}\right|\leq 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/886456437373f1c716a46a5e28c86750db91b79f)
Another corollary is that
converges whenever
is a decreasing sequence that tends to zero. To see that
is bounded, we can use the summation formula[2]
![{\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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3241d8c48ef36b9df294ddeae463f59a16fa886)
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
- ^ 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 [1].
- ^ "Where does the sum of $\sin(n)$ formula come from?".
References
- Hardy, G. H., A Course of Pure Mathematics, Ninth edition, Cambridge University Press, 1946. (pp. 379–380).
- Voxman, William L., Advanced Calculus: An Introduction to Modern Analysis, Marcel Dekker, Inc., New York, 1981. (§8.B.13–15) ISBN 0-8247-6949-X.
External links