I was reading a proof of Gauss' test here , but I have a problem in the last case:
As I know the author has used Kummer's test which is as follows:
Let $\left\{a_{n}\right\}_{n\ge N}\ ,\left\{b_{n}\right\}_{n\ge N}$ be two positive real sequences , define: $$c_{n}:=b_{n}\frac{a_{n}}{a_{n+1}}-b_{n+1}$$
Then:
- $\sum_{n=1}^{\infty}a_{n}$ converges if there exist $r>0$ and positive natural number $N$ such that $r\le c_{n}$ for all $n\ge N$
- $\sum_{n=1}^{\infty}a_{n}$ diverges if $c_{n}\le0$ for all $n\ge N$ and $\sum_{n=N}^{\infty}\frac{1}{b_{n}}$ diverges.
The problem I have is that the $c_n$ defined in Kummer's test is different from what is used at that site, so I don't know how to continue, can someone please explain why these two are different?(also does there exist a way to prove the last case where $A=1$ using Kummer's definition for $c_n$?)
To begin, it seems that we are only interested in the case of divergence so only need to look at the second part of the statement.
Now, let $\{a_n\}$ and $\{b_n\}$ be as in the theorem. Now, have $c_n:=b_{n}\frac{a_{n}}{a_{n+1}}-b_{n+1}$ and $c'_n= b_n-\frac{a_{n+1}b_{n+1}}{a_n}$ (the version from the proof). Suppose that $c_n\leq 0$ iff $c'_n\leq 0$. That would mean that the Kummer test is equivalent to a new test, where we just replace $c$ with $c'$, because then $c'_n\leq 0$ would imply $c_n\leq 0$. However, $c'_n = \frac{a_{n+1}}{a_n}c_n$, and $a_n$ is a positive sequence. That means that $c'_n$ and and $c_n$ have the same sign.
In summary, the reason the different $c_n$s are different is because all we care about is the sign. As long as we are multiplying by a positive number, we still get something with the same sign and so the test still holds. The proof chooses to do this in order to make some of the computation nicer. However, you could go back to the original using that Kummer test's $c_n$ is just $\frac{a_{n}}{a_{n+1}}$ times the other version of $c_n$. The proof would be the same, after showing that $c'_n\leq 0$, you would simply multiply by $\frac{a_{n}}{a_{n+1}}$ to recover $c_n$, and apply the Kummer test without modification.