Abel's Theorem

Theorem
Let $$\sum{x_{n}}$$ be a real terms series and let $$a_{n}$$ be a sequence of real elements. Abel's Theorem states that: if the sequence $$a_{n}$$ is monotone and it's bounded and if the series $$\sum{x_{n}}$$ converges, then the series $$\sum{a_nx_{n}}$$ is convergent.

Proof
First we need the following lemma.

Lemma
Let $$a_{n}$$ be a monotone and bounded sequence and let $$s_{n}$$ be a bounded sequence; put $$u_{n}=\left( a_{n}-a_{n+1} \right)s_{n},$$ then the series $$\sum{u_{n}}$$ is absolutely convergent.

Proof of Lemma
Since $$s_{n}$$ is bounded, $$\exist k>0$$ such that $$|s_n|\le k$$ for $$n\in\mathbb N.$$

Now put $$\sum\limits_{i=1}^{n }{\left| u_{i} \right|}=\sum\limits_{i=1}^{n }{\left| a_{i}-a_{i+1} \right|\left| s_{i} \right|}\le k\sum\limits_{i=1}^{n }{\left| a_{i}-a_{i+1} \right|}$$.

Since $$a_{n}$$ is a monotone sequence, it verifies that $$\sum\limits_{i=1}^{n}{\left| a_{i}-a_{i+1} \right|}=\left| a_{1}-a_{n+1} \right|\le \left| a_{1}-a \right|$$, where $$a=\underset{n\to \infty }{\mathop{\lim }}\,a_{n}$$.

Note: observe that $$a$$ exists and it's finite since $$a_{n}$$ is monotone bounded.

Then $$\sum\limits_{i=1}^{n}{\left| u_{i} \right|}\le k\left| a_{1}-a \right|$$, which implies that the series $$\sum{|u_{n}|}$$ has its partial sums bounded, hence $$\sum{|u_{n}|}$$ converges.

Therefore $$\sum{u_{n}}$$ is absolutely convergent.

Q.E.D.

Proof of Abel's Theorem
Now we're ready to prove Abel's Theorem:

Let $$s_n$$ be the $$n$$th sum of $$\sum{x_{n}}$$.

Also, let $$u_{n}=\left( a_{n}-a_{n+1} \right)s_{n},$$ then the series $$\sum{u_{n}}$$ satisfies the above Lemma.

On the other hand we have that $$\underset{n\to \infty }{\mathop{\lim }}\,a_{n}s_{n}=a\cdot s$$, where $$s=\underset{n\to \infty }{\mathop{\lim }}\,s_{n}$$ does exist and it's finite since the sum of the series $$\sum{x_{n}}$$ converges.

Finally, $$\sum\limits_{n=1}^{\infty }{a_{n}x_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{a_{i}x_{i}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n-1}{u_{i}}+\underset{n\to \infty }{\mathop{\lim }}\,a_{n}s_{n}=l+a\cdot s\in \mathbb{R},$$ which proves the convergence of $$\sum{a_nx_{n}}.$$

Q.E.D.