User:PAR/Test2

Extension to Kummer's test

Assuming:

• a_n > 0 for all n
• D_n > 0 for all n

Kummer's test defines:

$\rho _{n}=\left(D_{n}-D_{n+1}{\frac {a_{n+1}}{a_{n}}}\right)$

and states that:

• If there exists a c such that $\rho _{n}\geq c>0$ for all n, then $\sum _{k=1}^{\infty }a_{k}$ converges
• If $\rho _{n}\leq 0$ for all n, and $\sum _{k=1}^{\infty }1/D_{k}$ diverges, then $\sum _{k=1}^{\infty }a_{k}$ diverges.

The proposed extension is written:

${\overline {\rho }}_{n}=\left(D_{n}(1+\epsilon _{n})-D_{n+1}{\frac {a_{n+1}}{a_{n}}}\right)$

where ε_n is a sequence of real numbers such that $\lim _{n\to \infty }D_{n}\epsilon _{n}=0$.

I think it can be shown that:

• If there exists a c such that ${\overline {\rho }}_{n}\geq c>0$ for all n, then there exists a c such that $\rho _{n}\geq c>0$ for all n and so $\sum _{k=1}^{\infty }a_{k}$ converges.
• If ${\overline {\rho }}_{n}<0$ for all n, and $\sum _{k=1}^{\infty }1/D_{k}$ diverges, then $\rho _{n}<0$ for all n and so $\sum _{k=1}^{\infty }a_{k}$ diverges.

The problem remains to either show that when ${\overline {\rho }}_{n}=0$ and $\sum _{k=1}^{\infty }1/D_{k}$ diverges, $\sum _{k=1}^{\infty }a_{k}$ diverges, or to find further restrictions on ε_n such that it does diverge, so that the convergence properties specified by ${\overline {\rho }}_{n}$ are the same as specified by $\rho _{n}$.

The above equation for ${\overline {\rho }}_{n}=0$ can be solved for a_n:

$a_{n}=a_{1}D_{1}{\frac {Q_{n}}{D_{n}}}$

where Q_n is the product:

$Q_{n}=\Pi _{k=1}^{n-1}(1+\epsilon _{k})$

and Q₁=1. We may take a_n and D_n to be unity, without loss of generality. So now we wish to find when

$\sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }{\frac {Q_{k}}{D_{k}}}$

diverges. (Note that when ε_n = 0, the product Q_n = 1, and we recover the simple Kummer's test, and so $\sum _{k=1}^{\infty }a_{k}$ diverges since $\sum _{k=1}^{\infty }1/D_{k}$ diverges).

Abel's test applied to the above sequences states that if:

• $\sum _{k=1}^{\infty }\epsilon _{k}Q_{k}$ converges
• $1/\epsilon _{n}D_{n}$ is monotone and bounded

then

$\sum _{k=1}^{\infty }{\frac {\epsilon _{k}Q_{k}}{\epsilon _{k}D_{k}}}=\sum _{k=1}^{\infty }a_{k}$

coverges. In the present case, $1/\epsilon _{n}D_{n}$ is NOT bounded since ε_nD_n -> 0, but I think Abel's theorem can be modified to prove that if

• $\sum _{k=1}^{\infty }\epsilon _{k}Q_{k}$ converges
• $1/\epsilon _{n}D_{n}$ is unbounded

then

$\sum _{k=1}^{\infty }{\frac {\epsilon _{k}Q_{k}}{\epsilon _{k}D_{k}}}=\sum _{k=1}^{\infty }a_{k}$

diverges.

So now we just need to deal with the first assumption ($\sum _{k=1}^{\infty }\epsilon _{k}Q_{k}$ converges). It can be easily seen that

$Q_{n+1}-Q_{n}=(1+\epsilon _{n})Q_{n}-Q_{n}=\epsilon _{n}Q_{n}$

so we want to look at the convergence of $\sum _{k=1}^{\infty }(Q_{k+1}-Q_{k})$ which equals Q-1 where Q is defined as:

$Q=\lim _{n\to \infty }\Pi _{k=1}^{n-1}(1+\epsilon _{k})$

SO THE CONDITION THAT Q EXISTS is the restriction the ε_n must obey in order that $\sum _{k=1}^{\infty }a_{k}$ diverge for the case when ${\overline {\rho }}_{n}=0$.

According to Knopp (page 224, Theorem 9), if $\sum _{k=1}^{\infty }|\epsilon _{k}|$ converges, Q_n will converge to Q. Also according to Knopp (page 225 supplementary theorem), if $\sum _{k=1}^{\infty }\epsilon _{k}^{2}$ converges, Q_n will converge to Q.

So these are two restrictions on ε_n that will assure divergence of $\sum _{k=1}^{\infty }a_{k}$: ε_nD_n must converge to zero, and one or both of the above two conditions on ε_n apply. (Note these two conditions are sufficient, but not necessary: if they don't apply, that doesn't mean Q does not exist)

Using the extensions to prove extensions to Raabe's and Bertrand's tests and Gauss's test

• For Raabe's test, use D_n = n and $\epsilon _{n}={\frac {B_{n}}{n^{2}}}$ where B_n is a bounded sequence. All of the above requirement on ε_n are met.

• For Gauss's test, use D_n = n and $\epsilon _{n}={\frac {B_{n}}{n^{r}}}$ where B_n is a bounded sequence and r > 1. All of the above requirement on ε_n are met.

• For Bertrand's test, use D_n = n ln(n) and $\epsilon _{n}={\frac {B_{n}}{n\ln(n)^{2}}}$ where B_n is a bounded sequence. All of the above requirement on ε_n are met.

Since the ratio test is Kummer's test for D_n = 1, similar extensions could be made to the ratio test.