In calculus, the Abel–Dini–Pringsheim theorem is a convergence test which constructs from a divergent series a series that diverges more slowly, and from convergent series one that converges more slowly.[1]: §IX.39 Consequently, for every convergence test based on a particular series there is a series about which the test is inconclusive.[1]: 299 For example, the Raabe test is essentially a comparison test based on the family of series whose
th term is
(with
) and is therefore inconclusive about the series of terms
which diverges more slowly than the harmonic series.
The Abel–Dini–Pringsheim theorem can be given for divergent series or convergent series. Helpfully, these definitions are equivalent, and it suffices to prove only one case. This is because applying the Abel–Dini–Pringsheim theorem for divergent series to the series with partial sum
![{\displaystyle S_{n}'={\frac {1}{r_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2adf887397265f2e632bc182dcdc8d771e26a537)
yields the Abel–Dini–Pringsheim theorem for convergent series.[2]
For divergent series
[edit]
Suppose that
is a sequence of positive real numbers such that the series
![{\displaystyle \sum _{n=0}^{\infty }a_{n}=\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa18babadb12047e7f0618e3ceec3add23209755)
diverges to infinity. Let
denote the
th partial sum. The Abel–Dini–Pringsheim theorem for divergent series states that the following conditions hold.
![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}}}=\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/026b4445e43e0c45caa140fd8206982ae78a0cdb)
- For all
we have ![{\displaystyle \sum _{n=1}^{\infty }{\frac {a_{n}}{S_{n}S_{n-1}^{\epsilon }}}<\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cce94f73dc9ca0500fdd6a6299b10f86c54c060d)
- If also
, then ![{\displaystyle \lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{n}}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7eae26ab8eccfa8a4da2999f37085343152446ca)
Consequently, the series
![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}^{t}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a85863b0310915c506e9a06e6f521a89ae3ebbc)
converges if
and diverges if
. When
, this series diverges less rapidy than
.[1]
Proof
Proof of the first part. By the assumption
is nondecreasing and diverges to infinity. So, for all
there is
such that
![{\displaystyle {\frac {S_{n}}{S_{n+k_{n}}}}<{\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53f0f8938c109d1049aaef9e25198e765ea0c50c)
Therefore
![{\displaystyle {\frac {a_{n+1}}{S_{n+1}}}+\cdots +{\frac {a_{n+k_{n}}}{S_{n+k_{n}}}}\geq {\frac {a_{n+1}+\cdots +a_{n+k_{n}}}{S_{n+k_{n}}}}={\frac {S_{n+k_{n}}-S_{n}}{S_{n+k_{n}}}}=1-{\frac {S_{n}}{S_{n+k_{n}}}}>{\frac {1}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db8490363bb056aca9168cb66be46527708ac43b)
and hence
is not a Cauchy sequence. This implies that the series
![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1057e6e666863e10db19d5e04293f8bce7dd117c)
is divergent.
Proof of the second part. If
, we have
for sufficiently large
and thus
. So, it suffices to consider the case
. For all
we have the inequality
![{\displaystyle \epsilon (1-x)\leq 1-x^{\epsilon }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2f0f6478af15f7df0a4876a724daab45404e65b)
This is because, letting
![{\displaystyle f(x)=\epsilon (1-x)-1+x^{\epsilon }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b273b6fdb09f5440e8b6b7202c8cad441832874)
we have
![{\displaystyle f(1)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da761295ab324acdb04e33ab75299c7fb588675f)
![{\displaystyle f'(x)=\epsilon (x^{\epsilon -1}-1)\geq 0\qquad (\forall x\in (0,1])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f50b494962837929eda755819a9605f159860c8)
![{\displaystyle f'(x)=\epsilon (x^{\epsilon -1}-1)\leq 0\qquad (\forall x\in [1,\infty )).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6620f6696fb30e0c34d459b28bb7d6c789db6e11)
(Alternatively,
is convex and its tangent at
is
)
Therefore,
![{\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {a_{n}}{S_{n}S_{n-1}^{\epsilon }}}&=\sum _{n=1}^{\infty }{\frac {S_{n}-S_{n-1}}{S_{n}S_{n-1}^{\epsilon }}}\\&=\sum _{n=1}^{\infty }{\frac {1}{S_{n-1}^{\epsilon }}}\left(1-{\frac {S_{n-1}}{S_{n}}}\right)\\&\leq \sum _{n=1}^{\infty }{\frac {1}{\epsilon S_{n-1}^{\epsilon }}}\left(1-\left({\frac {S_{n-1}}{S_{n}}}\right)^{\epsilon }\right)\\&=\sum _{n=1}^{\infty }{\frac {1}{\epsilon }}\left({\frac {1}{S_{n-1}^{\epsilon }}}-{\frac {1}{S_{n}^{\epsilon }}}\right)\\&={\frac {1}{\epsilon S_{0}^{\epsilon }}}\\&<\infty \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6619b581d264a448cc3fc66338ca18104e94d72)
Proof of the third part. The sequence
is nondecreasing and diverges to infinity. By the Stolz-Cesaro theorem,
![{\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{n}}}&=\lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{0}+\ln(S_{1}/S_{0})+\cdots +\ln(S_{n}/S_{n-1})}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(S_{n}/S_{n-1})}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(S_{n}/(S_{n}-a_{n}))}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(1/(1-a_{n}/S_{n}))}}\\&=\lim _{n\to \infty }\left(-{\frac {a_{n}/S_{n}}{\ln(1-a_{n}/S_{n})}}\right)\\&=1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/415daeed0fe96d6a0a2a103deb6a02b90b6b1f4a)
For convergent series
[edit]
Suppose that
is a sequence of positive real numbers such that the series
![{\displaystyle \sum _{n=0}^{\infty }a_{n}<\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d68330737e6080063362103d972c4635a80dc13c)
converges to a finite number. Let
denote the
th remainder of the series. According to the Abel–Dini–Pringsheim theorem for convergent series, the following conditions hold.
![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}}}=\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/409cc2dd223770959afb013f180d1f6696cd6747)
- For all
we have ![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}^{1-\epsilon }}}<\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c284566d40a2050e797490077d94b5054e8807a8)
- If also
then ![{\displaystyle \lim _{n\to \infty }{\frac {a_{0}/r_{0}+a_{1}/r_{1}+\cdots +a_{n}/r_{n}}{\ln r_{n}}}=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c62558aba4612b5b3114341089b0eaef3fa932f7)
In particular, the series
![{\displaystyle \sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}^{t}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d66f080dfa961e6fddb8e71a6303de49ed4778ff)
is convergent when
, and divergent when
. When
, this series converges more slowly than
.[1]
The series
![{\displaystyle \sum _{n=0}^{\infty }1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/058c2eb0c4f215b55108cc2c9530b2b4150ac558)
is divergent with the
th partial sum being
. By the Abel–Dini–Pringsheim theorem, the series
![{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{t}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c033bc463fee00773b0524e34852955a973c1328)
converges when
and diverges when
. Since
converges to 0, we have the asymptotic approximation
![{\displaystyle \lim _{n\to \infty }{\frac {1+1/2+\cdots +1/n}{\ln n}}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/166fef61db6a344051857e5fc56da1a71c6c7357)
Now, consider the divergent series
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb88a65195479de89de202648fb46f7ddb89d475)
thus found. Apply the Abel–Dini–Pringsheim theorem but with partial sum replaced by asymptotically equivalent sequence
. (It is not hard to verify that this can always be done.) Then we may conclude that the series
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n\ln ^{t}n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/961994bcb19b1705265c7e65b5529a52d1dda3ab)
converges when
and diverges when
. Since
converges to 0, we have
![{\displaystyle \lim _{n\to \infty }{\frac {1+1/(2\ln 2)+\cdots +1/(n\ln n)}{\ln \ln n}}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf9705131f7d040699da31b4eb3a856d04cea5f6)
The theorem was proved in three parts. Niels Henrik Abel proved a weak form of the first part of the theorem (for divergent series).[3] Ulisse Dini proved the complete form and a weak form of the second part.[4] Alfred Pringsheim proved the second part of the theorem.[5] The third part is due to Ernesto Cesàro.[6]