User:PAR/Test2

Extension to Kummer's test
Assuming:
 * an > 0 for all n
 * Dn > 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 $$\rho_n \ge c > 0$$ for all n, then $$\sum_{k=1}^\infty a_k$$ converges
 * If $$\rho_n \le 0$$ 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 &epsilon;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 $$\overline{\rho}_n \ge c > 0$$ for all n, then there exists a c such that $$\rho_n \ge c > 0$$ 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 &epsilon;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 an:



a_n = a_1 D_1 \frac{Q_n}{D_n} $$

where Qn is the product:



Q_n = \Pi_{k=1}^{n-1}(1+\epsilon_k) $$

and Q1=1. We may take an and Dn 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 &epsilon;n = 0, the product Qn = 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: then
 * $$\sum_{k=1}^\infty \epsilon_k Q_k$$ converges
 * $$1/\epsilon_n D_n$$ is monotone and bounded
 * $$\sum_{k=1}^\infty \frac{\epsilon_k Q_k}{\epsilon_k D_k} = \sum_{k=1}^\infty a_k$$

coverges. In the present case, $$1/\epsilon_n D_n$$ is NOT bounded since &epsilon;nDn -> 0, but I think Abel's theorem can be modified to prove that if then
 * $$\sum_{k=1}^\infty \epsilon_k Q_k$$ converges
 * $$1/\epsilon_n D_n$$ is unbounded
 * $$\sum_{k=1}^\infty \frac{\epsilon_k Q_k}{\epsilon_k D_k} = \sum_{k=1}^\infty a_k$$

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 &epsilon;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, Qn will converge to Q. Also according to Knopp (page 225 supplementary theorem), if $$\sum_{k=1}^\infty \epsilon_k^2$$ converges, Qn will converge to Q.

So these are two restrictions on &epsilon;n that will assure divergence of $$\sum_{k=1}^\infty a_k$$:  &epsilon;nDn must converge to zero, and one or both of the above two conditions on &epsilon;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 Dn = n and $$\epsilon_n=\frac{B_n}{n^2}$$ where Bn is a bounded sequence. All of the above requirement on &epsilon;n are met.


 * For Gauss's test, use Dn = n and $$\epsilon_n=\frac{B_n}{n^r}$$ where Bn is a bounded sequence and r > 1. All of the above requirement on &epsilon;n are met.


 * For Bertrand's test, use Dn = n ln(n) and $$\epsilon_n=\frac{B_n}{n \ln(n)^2}$$ where Bn is a bounded sequence. All of the above requirement on &epsilon;n are met.

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