# Kummer's test - Calculus, Apostol, 10.16 #15.

I want to prove the following:

Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences with $a_n>0$ and $b_n>0$ for all $n \geq N$, and let $$c_n = b_n - \frac{a_{n+1}b_{n+1}}{a_n}$$

Then

1. If there exists $r>0$ such that $0<r\leq c_n \text{ ; } \forall n\geq N$ then $\displaystyle\sum a_n$ converges.
2. If $c_n\leq0$ for $n\geq N$ and if $\displaystyle\sum \dfrac{1}{b_n}$ diverges, then so does $\displaystyle\sum a_n$.

So far I know how to use and prove Cauchy's, D'Alambert's and the integral criterion, so I'd like a hint on either using those or a new idea. The book suggests that for $1$, I show that

$$\sum\limits_{k = N}^n {{a_k} \leq \frac{{{a_N}{b_N}}}{r}}$$

and for $2$, to prove that $\displaystyle\sum a_n$ dominates $\displaystyle\sum \dfrac{1}{b_n}$

I'd like to prove this with previous theory on series convergence (Cauchy's and/or D'Alambert's preferrably, comparison test) since it is what precedes the problem, and would appreaciate great HINTS rather than answers.

I can't seem to understand the inequalities. I mean, Cauchy's and D'Alambert's criteria reveal that the proof relies on the fact that a geometric series of with ratio $|r| < 1$ converges, but I can't understand the motivation in this proof.

Summing up the work.

For 1:

\eqalign{ & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{b_{N + 1}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{b_{N + 2}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \frac{{{a_{N + 3}}}}{{{a_N}}}{b_{N + 3}} \cr}

Induction over $n$ we get

$${b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}{c_{N + n}} + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}$$

So

\eqalign{ & {b_N} \geqslant r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}} \cr & {b_N} > r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) \cr & \frac{{{a_N}{b_N}}}{r} > {a_N} + {a_{N + 1}} + {a_{N + 2}} + \cdots + {a_{N + n}} = \sum\limits_{k = N}^n {{a_k}} \cr}

As desired.

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This appears to be known as "Kummer's test", by the way. – Dylan Moreland Mar 1 '12 at 4:35
@DylanMoreland Good to know. With this one can prove Raabe's test, and the prove Gauss's test. That's why it's so important to me. – Pedro Tamaroff Mar 1 '12 at 4:40

## 3 Answers

Here’s a hint for (1):

\begin{align*}b_N &= c_N + \frac{a_{N+1}b_{N+1}}{a_N}\\ &=c_N+\frac{a_{N+1}}{a_N}\left(c_{N+1}+\frac{a_{N+2}b_{N+2}}{a_{N+1}}\right)\\ &=c_N+\frac{a_{N+1}}{a_N}\cdot c_{N+1}+\frac{a_{N+2}}{a_N}\cdot b_{N+2}\\ &=c_N+\frac{a_{N+1}}{a_N}\cdot c_{N+1}+\frac{a_{N+2}}{a_N}\cdot c_{N+2}+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\\ &\ge r\left(1+\frac{a_{N+1}}{a_N}+\frac{a_{N+2}}{a_N}\right)+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\tag{why?}\\ &=\frac{r}{a_N}(a_N+a_{N+1}+a_{N+2})+\frac{a_{N+3}}{a_N}\cdot b_{N+3}\\ &>\frac{r}{a_N}(a_N+a_{N+1}+a_{N+2})\;.\tag{why?} \end{align*}

Generalize to get an upper bound on the partial sums of $\sum a_n$.

Added: A hint for (2): if $$b_n\le\frac{a_{n+1}b_{n+1}}{a_n}\;,$$ then $$\frac{a_n}{1/b_n}=a_nb_n\le a_{n+1}b_{n+1}=\frac{a_{n+1}}{1/b_{n+1}}\;.$$

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I see. That will do the work! I'll finish it so I/you can focus on $2$. Thanks! – Pedro Tamaroff Mar 1 '12 at 5:18
@Peter: (2) took me a little while, but I think that I’ve come up with a usable hint that doesn’t just give it away outright. – Brian M. Scott Mar 1 '12 at 5:29

Since $a_{n} > 0$ and $c_{n}\ge r$, we have $$a_{k}b_{k} - a_{k+1}b_{k+1} \ge r a_{k}.$$ Summing up from $k=N$ to $k=n$, $$\sum_{k=N}^{n}\left(a_{k}b_{k}-a_{k+1}b_{k+1}\right) \ge r\sum_{k=N}^{n}a_{k}.$$ The LHS is a telescopic sum, yielding $$\sum_{k=N}^{n}a_{k} \le \frac{a_{N}b_{N} - a_{n+1}b_{n+1}}{r}.$$ Since $a_n, b_n >0$, we get the desired upper bound: $$\sum_{k=N}^{n}a_{k}\le\frac{a_{N}b_{N}}{r}.$$

The second inequality may be deduced by the same fashion.

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Look at the expression for $c_n$ again. Doesn't the way the $a_n$ and the $a_{n+1}$ are arranged remind you of a particular test? Try to make certain assumptions about $\{c_n\}$ and see if you can then set up some inequalities of your own. Also, the inequality in 1. simply says $\forall n \geq N,\ c_n > 0.$ Keeping this in mind, enumerate the various possibilities for the ratio $a_n \over a_{n+1}$ and $b_n$ and $b_{n+1}$.

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Yes. I'm thinking about D'Alambert but I can't figure it out yet. – Pedro Tamaroff Mar 1 '12 at 4:42
@PeterT.off Ok, the inequality in 1. does give away something important. I'll just edit the answer. – ThisIsNotAnId Mar 1 '12 at 4:59