# Proving Abel's test of convergence

I am self-learning Real Analysis from the text, Understanding Analysis, by Stephen Abbott. I would like to prove Abel's test of convergence in the following exercise problem 2.7.12. A clue/hint, without revealing the entire solution would be extremely helpful.

I reproduce my attempt for part(a) of the problem below.

(Abel's Test). Abel's test for convergence states that if the series $$\displaystyle \sum _{k=1}^{\infty } x_{k}$$ converges and if $$\displaystyle ( y_{k})$$ is a sequence satisfying:

$$\begin{equation*} y_{1} \geq y_{2} \geq y_{3} \geq \dotsc \geq 0 \end{equation*}$$ then the series $$\displaystyle \sum _{k=1}^{\infty } x_{k} y_{k}$$ converges.

(a) Use exercise 2.7.12 to show that:

$$\begin{equation*} \sum _{k=1}^{n} x_{k} y_{k} =s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1}) \end{equation*}$$

where $$\displaystyle s_{n} =x_{1} +\dotsc +x_{n}$$.

Proof.

By the formula for summation by parts, we have:

$$\begin{equation*} \begin{array}{ c l } \sum _{k=1}^{n} x_{k} y_{k} & =s_{n} y_{n+1} -s_{0} y_{1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\\ & =s_{n} y_{n+1} -0\cdot y_{1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\\ & =s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1}) \end{array} \end{equation*}$$

(b) Use the comparison test to argue that: $$\displaystyle \sum _{k=1}^{\infty } s_{k}( y_{k} -y_{k+1})$$ converges absolutely, and show how this leads directly to a proof of Abel's test.

• Hint: $\left(s_n\right)$ is bounded (because it converges) and $\left(\sum \left(y_n - y_{n+1}\right)\right)$ is a telescoping series. Nov 20, 2022 at 10:41
• $\{y_n\}$ is i.e. any non-increasing non-negative sequence?
– BCLC
Nov 25, 2022 at 17:36
• @BCLC, yup, that's what the statement in my book says. Abbott Page 79. Nov 25, 2022 at 18:03

My proof attempt for the sake of completeness.

Proof.

Since $$\displaystyle \sum _{k=1}^{\infty } x_{k}$$ converges, $$\displaystyle ( s_{k})$$ is a convergent sequence and hence it is bounded. There exists $$\displaystyle M >0$$ for all $$\displaystyle k\in \mathbf{N}$$, such that $$\displaystyle |s_{k} |\leq M$$. Thus,

$$\begin{equation*} |s_{k}( y_{k} -y_{k+1}) |\leq M|y_{k} -y_{k+1} |\leq M( y_{k} -y_{k+1}) \quad \{\because y_{k} \geq y_{k+1}\} \end{equation*}$$

We know that $$\displaystyle ( y_{k} -y_{k+1}) \geq 0$$. Let $$\displaystyle ( t_{k})$$ be the sequence of partial sums of the infinite series $$\displaystyle \sum _{k=1}^{\infty }( y_{k} -y_{k+1})$$. Since $$\displaystyle ( t_{k})$$ is a monotonically-increasing sequence and

$$\begin{equation*} \begin{array}{ c l } t_{k} & =( y_{1} -y_{2}) +( y_{2} -y_{3}) +\dotsc +( y_{k} -y_{k+1})\\ & =y_{1} -y_{2} +y_{2} -y_{3} +y_{3} +\dotsc +y_{k} -y_{k+1}\\ & =y_{1} -y_{k+1}\\ & \leq y_{1} \end{array} \end{equation*}$$

it is bounded by $$\displaystyle y_{1}$$, by the Montone convergence theorem, $$\displaystyle t_{k}$$ is a convergent series. By the Algebraic limit theorem for infinite series, $$\displaystyle M\cdot \sum _{k=1}^{\infty }( y_{k} -y_{k+1})$$ is a convergent series.

Hence, by the comparison test, $$\displaystyle \sum _{k=1}^{\infty } |s_{k}( y_{k} -y_{k+1}) |$$ is a convergent series.

By the Absolute convergence test, $$\displaystyle \sum _{k=1}^{\infty } s_{k}( y_{k} -y_{k+1})$$ is a convergent series.

Passing to the limits, we have:

$$\begin{equation*} \begin{array}{ c c } \lim _{n\rightarrow \infty }\sum _{k=1}^{n} x_{k} y_{k} & =\lim _{n\rightarrow \infty }\left[ s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\right] \end{array} \end{equation*}$$

Note that, $$\displaystyle ( y_{n})$$ is bounded by $$\displaystyle [ 0,y_{1}]$$ and is a monotonically decreasing sequence. Hence, by MCT, it is a convergent sequence. $$\displaystyle ( s_{n})$$ is also given to be a convergent sequence. Hence, the limit of the right hand side of the expression can be written as:

$$\begin{equation*} \lim _{n\rightarrow \infty }\left[ s_{n} y_{n+1} +\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1})\right] =\lim _{n\rightarrow \infty } s_{n} y_{n+1} +\lim _{n\rightarrow \infty }\sum _{k=1}^{n} s_{k}( y_{k} -y_{k+1}) \end{equation*}$$

Since both these limits exist, the limit on the right hand side exists. Thus, the product series $$\displaystyle \sum _{k=1}^{\infty } x_{k} y_{k}$$ converges.