Proof verification about Abel's test

Question

I've seen several proofs of Abel's test, but still I'm not sure if the proofs are as precious as they can be, hence I decided to proof the test and add some details whenever it's needed:

Theorem: Assume $\left\{a_{k}\right\}_{k\ge1},\left\{b_{k}\right\}_{k\ge1}$ are two real sequences, if $\sum_{k=1}^{\infty}a_{k}$ is convergent , $\left\{b_{k}\right\}_{k\ge1}$ is monotonic and $\sum_{k=1}^{\infty}b_{k}$ is convergent, then $\sum_{k=1}^{\infty}a_{k}b_{k}$ is also a convergent series.

Proof:

First define $$S_{n}:=\sum_{k=1}^{n}a_{k}\;\;\;\;,\;\;\text{and notice that} \;\;\;\;,\;\;S_0=0$$

and consider the following finite series:

$$\sum_{k=1}^{n}a_{k}b_{k}=\sum_{k=1}^{n}\left(S_{k}-S_{k-1}\right)b_{k}=\sum_{k=1}^{n}S_{k}b_{k}-\sum_{k=1}^{n}S_{k-1}b_{k}$$

Setting $k-1 \mapsto k$ in the right series follows:

$$=\sum_{k=1}^{n}S_{k}b_{k}-\sum_{k=0}^{n-1}S_{k}b_{k+1}=\sum_{k=1}^{n}S_{k}\left(b_{k}-b_{k+1}\right)+S_{n}b_{n+1}$$

Hence we showed that:

$$\bbox[5px,border:2px solid #00A000]{\sum_{k=1}^{n}a_{k}b_{k}=\sum_{k=1}^{n}S_{k}\left(b_{k}-b_{k+1}\right)+S_{n}b_{n+1}}$$

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Since $\sum_{k=1}^{\infty}a_{k}$ is convergent we conclude that $\lim_{n \to \infty}S_n$ does exist and is a finite value, from convergent sequences are bounded it follows that there exist a real number $M$ such that $\forall n \in \mathbb N^{+}$: $\left|S_{n}\right|\le M$, now it's the time to use this fact:

$$\sum_{k=1}^{n}\left|S_{k}\left(b_{k}-b_{k+1}\right)\right|=\sum_{k=1}^{n}\left|S_{k}\right|\left|\left(b_{k}-b_{k+1}\right)\right|\le\sum_{k=1}^{n}M\left|\left(b_{k}-b_{k+1}\right)\right|=M\sum_{k=1}^{n}\left|\left(b_{k}-b_{k+1}\right)\right|$$

If $\left\{b_{k}\right\}_{k\ge1}$ is monotone decreasing then we have: $$\sum_{k=1}^{n}\left|\left(b_{k}-b_{k+1}\right)\right|=\sum_{k=1}^{n}\left(b_{k}-b_{k+1}\right)=b_{1}-b_{n+1}$$

Taking the limit of this expression:

$$\sum_{k=1}^{\infty}\left|\left(b_{k}-b_{k+1}\right)\right|=\lim_{n \to \infty}b_{1}-\lim_{n \to \infty}b_{n+1}$$

From term test we conclude: $$\sum_{k=1}^{\infty}\left|\left(b_{k}-b_{k+1}\right)\right|=b_1$$

If $\left\{b_{k}\right\}_{k\ge1}$ is monotone increasing then we have:

$$\sum_{k=1}^{n}\left|\left(b_{k}-b_{k+1}\right)\right|=-\sum_{k=1}^{n}\left(b_{k}-b_{k+1}\right)=-b_{1}+b_{n+1}$$

Again using the previous result yields: $$\sum_{k=1}^{\infty}\left|\left(b_{k}-b_{k+1}\right)\right|=-b_1$$

In either cases it's seen that $\sum_{k=1}^{\infty}\left|\left(b_{k}-b_{k+1}\right)\right|$ is convergent, from Constant Multiple Rule of convergent series we see that $M\sum_{k=1}^{\infty}\left|\left(b_{k}-b_{k+1}\right)\right|$ converges and using comparison test tells us that $\sum_{k=1}^{\infty}\left|S_{k}\left(b_{k}-b_{k+1}\right)\right|$ is also convergent, on the other hand we know that Absolutely Convergent Series is Convergent, so $\sum_{k=1}^{\infty}S_{k}\left(b_{k}-b_{k+1}\right)$ is convergent, besides $S_{n}b_{n+1}$ is fixed value, concludes:

$$\sum_{k=1}^{\infty}a_{k}b_{k} \;\;\;\;\;\;\text{converges}$$

Q.E.D.

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I'm not sure if I've used the theorems properly, so can someone verify my proof?

Note

There is a little difference between this theorem and the theorem used in Wikipedia, because Wikipedia claims that the sequence $\left\{b_{k}\right\}_{k\ge1}$ must be bounded, but I replaced this condition to "$\sum_{k=1}^{\infty}b_{k}$ is convergent", if we just consider the Wiki's condition, then the proof won't change that much, I will show that my theorem implies Wiki's theorem but the converse is not true:

$\sum_{k=1}^{\infty}b_{k}$ is convergent then nth term test implies $$\lim_{k \to \infty}b_{k}=0 \iff \forall \epsilon >0, \exists N \in \mathbb N^{+}:\forall k\left(k\ge N \implies \left|b_{k}\right|<\epsilon\right)$$ So it's shown that the sequence $\left\{b_{k}\right\}_{k\ge N}$ is bounded.

Now I'm going to show that the converse is not true, a classic example is Harmonic series, because $\forall k \in \mathbb N^{+}$:

$$\frac{1}{k}\le1$$

However the sequence $\sum_{k=1}^{n}\frac{1}{k}$ is not convergent.

Answer 1

Your argument is sound (well done!) except for one detail at the end, where you say "besides $S_nb_{n+1}$ is fixed value". Since $S_nb_{n+1}$ depends on $n$, it does affect what happens when $n\to\infty$ in your green box. Fortunately the sequence $(S_n)$ is bounded, and by your hypothesis $b_n\to0$, so $\lim S_nb_{n+1}=0$, and you can properly conclude that $\sum a_kb_k$ converges.

Note that your theorem has the condition that $\sum b_k$ converges. This makes your theorem less general than the Wiki version. It is true that "$\sum b_k$ converges" implies "$(b_k)$ bounded", but this means that your condition is harder to satisfy, so your theorem is not usable in as many situations as the Wiki version. For example, you cannot apply your theorem to the case where $b_k$ is identically $1$ to conclude the obvious fact that $\sum a_k \cdot 1$ converges.