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 $C>0$.

Question

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.

Answer 1

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.

Answer 2

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.

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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.