Abel–Dini–Pringsheim theorem

In calculus, the Abel–Dini–Pringsheim theorem is a convergence test which constructs from a divergent series a series that diverges more slowly, or similarly for convergent series. Consequently, for every convergence test based on a particular series there is a series about which the test is inconclusive.

Definitions

For divergent series

Suppose that $(a_{n})_{n=0}^{\infty }\subset (0,\infty )$ is a sequence of positive real numbers such that the series

$\sum _{n=0}^{\infty }a_{n}=\infty$

diverges to infinity. Let $S_{n}=a_{0}+a_{1}+\cdots +a_{n}$ denote the $n$th partial sum. The Abel–Dini–Pringsheim theorem for divergent series states that the following conditions hold.

• $\sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}}}=\infty$
• For all $\epsilon >0$ we have $\sum _{n=1}^{\infty }{\frac {a_{n}}{S_{n}S_{n-1}^{\epsilon }}}<\infty$
• If also $\lim _{n\to \infty }{\frac {a_{n}}{S_{n}}}=0$, then $\lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{n}}}=1$

Consequently, the series

$\sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}^{t}}}$

converges if $t>1$ and diverges if $t\leq 1$.

Proof

Proof of the first part. By the assumption $S_{n}$ is nondecreasing and diverges to infinity. So, for all $n\in \{0,1,2,\dots \}$ there is $k_{n}\geq n+1$ such that

${\frac {S_{n}}{S_{n+k_{n}}}}<{\frac {1}{2}}$

Therefore

${\frac {a_{n+1}}{S_{n+1}}}+\cdots +{\frac {a_{n+k_{n}}}{S_{n+k_{n}}}}\geq {\frac {a_{n+1}+\cdots +a_{n+k_{n}}}{S_{n+k_{n}}}}={\frac {S_{n+k_{n}}-S_{n}}{S_{n+k_{n}}}}=1-{\frac {S_{n}}{S_{n+k_{n}}}}>{\frac {1}{2}}$

and hence $a_{0}/S_{0}+\cdots +a_{n}/S_{n}$ is not a Cauchy sequence. This implies that the series

$\sum _{n=0}^{\infty }{\frac {a_{n}}{S_{n}}}$

is divergent.

Proof of the second part. If $0<\epsilon \leq \epsilon '$, we have $S_{n}\geq 1$ for sufficiently large $n$ and thus $a_{n}/(S_{n}S_{n-1}^{\epsilon })\geq a_{n}/(S_{n}S_{n-1}^{\epsilon '})$. So, it suffices to consider the case $0<\epsilon \leq 1$. For all $x\in (0,\infty )$ we have the inequality

$\epsilon (1-x)\leq 1-x^{\epsilon }.$

This is because, letting

$f(x)=\epsilon (1-x)-1+x^{\epsilon }$

we have

$f(1)=0$

$f'(x)=\epsilon (x^{\epsilon -1}-1)\geq 0\qquad (\forall x\in (0,1])$

$f'(x)=\epsilon (x^{\epsilon -1}-1)\leq 0\qquad (\forall x\in [1,\infty )).$

(Alternatively, $g(x)=1-x^{\epsilon }$ is convex and its tangent at $1$ is $y=g'(1)(x-1)=\epsilon (1-x)$) Therefore,

${\begin{aligned}\sum _{n=1}^{\infty }{\frac {a_{n}}{S_{n}S_{n-1}^{\epsilon }}}&=\sum _{n=1}^{\infty }{\frac {S_{n}-S_{n-1}}{S_{n}S_{n-1}^{\epsilon }}}\\&=\sum _{n=1}^{\infty }{\frac {1}{S_{n-1}^{\epsilon }}}\left(1-{\frac {S_{n-1}}{S_{n}}}\right)\\&\leq \sum _{n=1}^{\infty }{\frac {1}{\epsilon S_{n-1}^{\epsilon }}}\left(1-\left({\frac {S_{n-1}}{S_{n}}}\right)^{\epsilon }\right)\\&=\sum _{n=1}^{\infty }{\frac {1}{\epsilon }}\left({\frac {1}{S_{n-1}^{\epsilon }}}-{\frac {1}{S_{n}^{\epsilon }}}\right)\\&={\frac {1}{\epsilon S_{0}^{\epsilon }}}\\&<\infty \end{aligned}}$

Proof of the third part. The sequence $\ln S_{n}$ is nondecreasing and diverges to infinity. By the Stolz-Cesaro theorem,

${\begin{aligned}\lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{n}}}&=\lim _{n\to \infty }{\frac {a_{0}/S_{0}+a_{1}/S_{1}+\cdots +a_{n}/S_{n}}{\ln S_{0}+\ln(S_{1}/S_{0})+\cdots +\ln(S_{n}/S_{n-1})}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(S_{n}/S_{n-1})}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(S_{n}/(S_{n}-a_{n}))}}\\&=\lim _{n\to \infty }{\frac {a_{n}/S_{n}}{\ln(1/(1-a_{n}/S_{n}))}}\\&=\lim _{n\to \infty }\left(-{\frac {a_{n}/S_{n}}{\ln(1-a_{n}/S_{n})}}\right)\\&=1\end{aligned}}$

For convergent series

Suppose that $(a_{n})_{n=0}^{\infty }\subset (0,\infty )$ is a sequence of positive real numbers such that the series

$\sum _{n=0}^{\infty }a_{n}<\infty$

converges to a finite number. Let $r_{n}=a_{n}+a_{n+1}+a_{n+2}+\cdots$ denote the $(n-1)$th remainder of the series. According to the Abel–Dini–Pringsheim theorem for convergent series, the following conditions hold.

• $\sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}}}=\infty$
• For all $\epsilon >0$ we have $\sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}^{1-\epsilon }}}<\infty$
• If also $\lim _{n\to \infty }{\frac {a_{n}}{r_{n}}}=0$ then $\lim _{n\to \infty }{\frac {a_{0}/r_{0}+a_{1}/r_{1}+\cdots +a_{n}/r_{n}}{\ln r_{n}}}=-1$

In particular, the series

$\sum _{n=0}^{\infty }{\frac {a_{n}}{r_{n}^{t}}}$

is convergent when $t<1$, and divergent when $t\geq 1$.

Equivalence

Applying the Abel–Dini–Pringsheim theorem for divergent series to the series with partial sum

$S_{n}'={\frac {1}{r_{n}}}$

yields the Abel–Dini–Pringsheim theorem for convergent series. Therefore, the two forms of the theorems are in fact equivalent.

Examples

The series

$\sum _{n=0}^{\infty }1$

is divergent with the $n$th partial sum being $n$. By the Abel–Dini–Pringsheim theorem, the series

$\sum _{n=0}^{\infty }{\frac {1}{n^{t}}}$

converges when $t>1$ and diverges when $t\leq 1$. Since $1/n$ converges to 0, we have the asymptotic approximation

$\lim _{n\to \infty }{\frac {1+1/2+\cdots +1/n}{\ln n}}=1.$

Now, consider the divergent series

$\sum _{n=1}^{\infty }{\frac {1}{n}}$

thus found. Apply the Abel–Dini–Pringsheim theorem but with partial sum replaced by asymptotically equivalent sequence $\ln n$. (It is not hard to verify that this can always be done.) Then we may conclude that the series

$\sum _{n=1}^{\infty }{\frac {1}{n\ln ^{t}n}}$

converges when $t>1$ and diverges when $t\leq 1$. Since $1/(n\ln n)$ converges to 0, we have

$\lim _{n\to \infty }{\frac {1+1/(2\ln 2)+\cdots +1/(n\ln n)}{\ln \ln n}}=1.$

History

Niels Henrik Abel proved a weak form of the first part of the theorem (for divergent series). Ulisse Dini proved the complete form and a weak form of the second part. Alfred Pringsheim proved the second part of the theorem. The third part is due to Ernesto Cesàro.