# 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.

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.

• If $a_k$ is a non-decreasing sequence of positive numbers, then $l=\lim\limits_{k\to\infty}a_k\ge a_1\gt0$.
– robjohn
Commented Aug 23, 2022 at 10:20
• @robjohn You are right. Thank you. Commented Aug 23, 2022 at 11:24
• @geetha290krm Sorry accidentally hit enter when I wanted to create a new line. I was also wondering how you would articulate that ln(1 + x) ~ x in a "epsilon-delta" proof. For example, Given epsilon > 0, there exists an N such that whenever n > m >= N, |b_{m+1} + b_{m+2} + ... + b_{n} | < epsilon and |b_{n}| < epsilon. I was hoping that this could lead to |ln(1-b_{m+1}) + ln(1-b_{m+2}) + ... + ln(1-b{n})| < epsilon. But I don't know error for ln(1 + x) ~ x, which I believe would be the first step toward making a claim like that. Commented Aug 25, 2022 at 1:20
• @person A precise argument (of a more general result) is available in Rudin's book. You can look for 'infinite prodiucts' in the index. Commented Aug 25, 2022 at 4:55

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.

• From (4) we conclude that if the sum converges then the product is bounded below by a positive number and must be positive, but how do we know that $1 - \sum_{k=1}^{\infty}b_k$ isn't negative? Commented Aug 25, 2022 at 2:19
• This is where the passage "Thus, if the sum converges, by removing a finite number of terms, we can assume that the partial sum is less than $\frac12$" comes into play. We can remove that finite product and see that the product of the terms whose sum is less than $\frac12$ gives a non-vanishing product. Then we can replace the finite product that we removed.
– robjohn
Commented Aug 25, 2022 at 3:41
• The statement at the beginning is not entirely precise. Take $b_1 =2$ and $b_n=n^{-2}$ for $n\geq 2$. The product is negative but the series still converge.
– Gary
Commented Aug 25, 2022 at 4:22
• @Gary: I have restricted the statement to match the question.
– robjohn
Commented Aug 25, 2022 at 5:16