Show that $\sum_{k=1}^{\infty}(l-a_k)$ converges if and only if $\lim_{n\to\infty}\frac{a_1\cdot a_2\cdot ...\cdot a_n}{l^n}=C$ for some $C>0$. Let $(a_n)_{n\geq1}$ be a nondecreasing sequence of positive numbers converging to some limit $l$.
Show that $\sum_{k=1}^{\infty}(l-a_k)$ converges if and only if $\lim_{n\to\infty}\frac{a_1\cdot a_2\cdot ...\cdot a_n}{l^n}=C$ for some $C>0$.
My thoughts:
In the "forward" direction, Given $\varepsilon>0$ there exists an $N$ such that whenever $m > n\geq N$,
$|\sum_{k=n+1}^{m}(1-\frac{a_k}{l})| < \frac{\varepsilon}{l}$ by the Cauchy Criterion.
We also know that $\lim\frac{a_n}{l}=1$ with $0<\frac{a_n}{l}<1$ for all $n$.
For finite $N$ we know that $(a_1a_2...a_N)/l^N$ is also positive and finite.
If anyone could offer any hints that'd be appreciated. Thanks.
 A: Note that $l\geq a_1>0$. Let $b_k=1-\frac {a_k} l$. The statement now becomes  the following:
$\sum b_k$ converges if and only if $\prod (1-b_k)$ converges to a positive number. This equivalence is a standard fact about infinite products. It is proved by going to logarithms and using that fact that $\ln (1+x) \sim x$ as $x \to 0$.  Ref: Rudin's RCA.
A: This answer is along similar lines as that of geetha290km, but here is a non-logarithmic proof that the sum and related product converge together.

Let $b_k=1-\frac{a_k}{l}$, then the question becomes

Let $(b_n)_{n\ge1}$ be a non-increasing sequence of numbers in $[0,1)$ converging to $0$.
Show that $\sum\limits_{k=1}^\infty b_k$ converges if and only if $\prod\limits_{k=1}^\infty(1-b_k)\gt0$.

Theorem: Suppose that $0\le b_k\lt1$, then
$$
\prod_{k=1}^n\frac1{1\mp b_k}\ge\prod_{k=1}^n(1\pm b_k)\ge1\pm\sum_{k=1}^nb_k\tag1
$$
Proof: The left-hand inequality of $(1)$ follows simply from $(1-x)(1+x)=1-x^2\le1$. We will show the right-hand inequality of $(1)$ by induction.
$(1)$ is trivial for $n=1$. Assume we have $(1)$ for $n-1$:
$$
\prod_{k=1}^{n-1}(1\pm b_k)\ge1\pm\sum_{k=1}^{n-1}b_k\tag2
$$
Then
$$
\begin{align}
\prod_{k=1}^n(1\pm b_k)
&=(1\pm b_n)\prod_{k=1}^{n-1}(1\pm b_k)\tag{3a}\\
&\ge(1\pm b_n)\left(1\pm \sum_{k=1}^{n-1}b_k\right)\tag{3b}\\
&=1\pm\sum_{k=1}^nb_k+b_n\sum_{k=1}^{n-1}b_k\tag{3c}\\
&\ge1\pm\sum_{k=1}^nb_k\tag{3d}
\end{align}
$$
Explanation:
$\text{(3a):}$ pull the $k=n$ term out front
$\text{(3b):}$ apply $(2)$
$\text{(3c):}$ expand the product
$\text{(3d):}$ $b_k\ge0$
Thus, we have $(1)$ for $n$.
$\large\square$
Convergence is dependent on what happens in the tail of the sequence of partial sums/products. Thus, if the sum converges, by removing a finite number of terms, we can assume that the partial sum is less than $\frac12$. Furthermore, if the product converges, by removing a finite number of terms, we can assume that the partial product of $1-b_k$ is between $\frac12$ and $1$ (and that the partial product of $1+b_k$ is between $1$ and $2$).
The Theorem says
$$
1-\sum_{k=1}^\infty b_k\le\prod_{k=1}^\infty(1-b_k)\tag4
$$
which says that if the sum converges, the product is bounded below, so it converges to a positive value.
Furthermore,
$$
1+\sum_{k=1}^\infty b_k\le\left(\prod_{k=1}^\infty(1-b_k)\right)^{-1}\tag5
$$
which says that if the product converges to a positive value, the sum is bounded above and so it converges.
