Prove $\sum_{n=1}^\infty a_n < \infty \implies \sum_{n=1}^\infty a_n(1-r^n) \to 0 \text{ as } r \to 1^-.$ Let $(a_n)$ be a sequence of positive terms and suppose $$\displaystyle\sum_{n=1}^\infty a_n < \infty.$$ Prove (from first principles) that $$\displaystyle\sum_{n=1}^\infty a_n(1-r^n) \to 0 \text{ as } r \to 1^-.$$
 A: This is famously known as Abel's theorem. We can phrase your statement as follows:

Theorem Suppose that $\displaystyle \sum_{n\geqslant 0}a_n$ converges, and define for $|x|<1$ the function $\displaystyle  f(x)=\sum_{n\geqslant 0}a_nx^n$ Then $\displaystyle  \lim_{x\to 1^{-}}f(x)=\sum_{n\geqslant 0}a_n$.

$(1)$ Denote by $A$ your sum. Since we're working in $|x|<1$, absolute convergence lets us write $$\begin{align}
  \frac{1}{{1 - x}}\left( {f\left( x \right) - A} \right) &= \frac{1}{{1 - x}}\sum\limits_{n \geqslant 0} {{a_n}{x^n}}  - \sum\limits_{n \geqslant 0} {A{x^n}}  \cr 
   &= \sum\limits_{n \geqslant 0} {{x^n}} \sum\limits_{n \geqslant 0} {{a_n}{x^n}}  \cr 
   &= \sum\limits_{n \geqslant 0} {\sum\limits_{k = 0}^n {{a_n}} {x^n}}  - \sum\limits_{n \geqslant 0} {A{x^n}} \cr 
   &= \sum\limits_{n \geqslant 0} \left(\sum\limits_{k = 0}^n {{a_n}} -A\right){x^n}   \end{align} $$
$(2)$ You know that given $\varepsilon$ you can make $$\left|{\sum\limits_{k = 0}^n {{a_n}}  - A}\right|<\varepsilon$$
and moreover we can assert there is a uniform bound $$\left| {\sum\limits_{k = 0}^n {{a_n}}  - A} \right| < M$$
$(3)$ Combine this and the fact that  
$$\displaylines{
  \sum\limits_{n \geqslant N} {{x^n}}  = \frac{{{x^N}}}{{1 - x}} \cr 
  \sum\limits_{n < N} {{x^n}}  = \frac{{1 - {x^N}}}{{1 - x}} \cr} $$
to obtain a really nice $x$ dependent upper bound of $|f(x)-A|$. Then, "let" $x\to 1^-$.

Spoiler ahead
Proof  Let $A$ be our sum. Observe that for $|x|<1$, we have that $$\frac{1}{1-x}f(x)=\sum_{n\geqslant 0}\sum_{k=0}^n a_kx^n$$
Thus, 

$$\frac{1}{1-x}\left( f(x)-A\right)=\sum_{n\geqslant 0}\left(\sum_{k=0}^n a_k-A\right)x^n$$

By hypothesis, there exists $N>0$ such that whenever $n\geqslant N$ we have 

$$\left|\sum_{k=0}^n a_k-A\right|<\varepsilon$$

Since we can assume $1-x>0$, we can write 

$$\begin{align}  \frac{1}{{1 - x}}\left| {f(x) - A} \right| &\leqslant \sum\limits_{n < N} {\left( {\sum\limits_{k = 0}^n {{a_k}}  - A} \right){x^n}}  + \sum\limits_{n \geqslant N} {\left( {\sum\limits_{k = 0}^n {{a_k}}  - A} \right){x^n}} \\  &\leqslant \sum\limits_{n < N} {\left( {\sum\limits_{k = 0}^n {{a_k}}  - A} \right){x^n}}  + \varepsilon \sum\limits_{n \geqslant N} {{x^n}} \\ &\leqslant \sum\limits_{n < N} {\left( {\sum\limits_{k = 0}^n {{a_k}}  - A} \right){x^n}}  + \varepsilon \frac{{{x^N}}}{{1 - x}} \end{align} $$

Now, let $M$ be a bound for the differences of partials sums, then we may write  

$$\frac{1}{{1 - x}}\left| {f(x) - A} \right| \leqslant M\frac{{1 - {x^N}}}{{1 - x}} + \varepsilon \frac{{{x^N}}}{{1 - x}}$$

Thus, we obtain 

$$\left| {f(x) - A} \right| \leqslant M\left( {1 - {x^N}} \right) + \varepsilon {x^N}$$ 

Now, we can choose $\delta$ such that whenever $1-x<\delta$, we have $1-x^N<\frac{\varepsilon}{2M}$, and since trivially $x^N<1$, we obtain for $x\in (1-\delta,1)$ that 

 $$\left| {f(x) - A} \right| \leqslant \frac{\varepsilon}2+\frac{\varepsilon}2<\varepsilon $$

$\blacktriangle$
A: Suppose $\displaystyle\sum_1^\infty a_n < \infty$ and fix $\varepsilon > 0$. Clearly $a_n (1-r^n) \leq a_n$ for all $r \in[0,1)$ and $n \geq 1$, so $\displaystyle\sum_1^\infty a_n (1-r^n) \leq \displaystyle\sum_1^\infty a_n < \infty$ for all $r \in [0,1)$. Moreover, since $\displaystyle\sum_1^\infty a_n$ converges, there exists $N \geq 1$ such that $\displaystyle\sum_N^\infty a_n (1-r^n) \leq \displaystyle\sum_N^\infty a_n < \frac{\varepsilon}{2}$ for all $r \in [0,1)$. Now, for all $n \in \{1,\ldots, N\}$ we have that $a_n r^n \to a_n$ as $r \to 1$ from below by continuity. That is, there exists $\delta_n \in (0,1)$ such that for all $r \in (\delta_n,1)$, $a_n - a_nr^n < \dfrac{\varepsilon}{2N}$. Choose $\delta = \displaystyle\max_{1 \leq n \leq N} \delta_n$. Then given $r \in (\delta,1)$, we have that $$\displaystyle\sum_1^N a_n(1-r^n) = \displaystyle\sum_1^N (a_n - a_nr^n) < \displaystyle\sum_1^N \frac{\varepsilon}{2N} = \frac{\varepsilon}{2}$$ and so that $$\sum_1^\infty a_n(1-r^n) = \sum_1^N a_n(1-r^n) + \sum_N^\infty a_n(1-r^n) <\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$
