# Other ways to show $y_n = \prod_{k=1}^n \left(1+{1\over x_k}\right)$ is bounded, where $x_n$ is another sequence with given properties.

Problem statement:

Let $$n\in \mathbb N$$ and $$\{x_n\}$$ be a sequence of natural numbers such that $$S_n$$ defined by: $$S_n = {1\over x_1} + {1\over x_2} + {1\over x_2} + \dots + {1\over x_n}$$ is bounded.

Show that $$y_n$$ is bounded given: $$y_n = \prod_{k=1}^n\left(1 + {1\over x_k}\right)$$

From the problem statement we know that $$S_n$$ is bounded. Let's try to take log of $$y_n$$: $$\ln y_n = \ln\left(\prod_{k=1}^n\left(1 + {1\over x_k}\right)\right) = \ln\left(1 + {1\over x_1}\right) + \ln\left(1 + {1\over x_2}\right) + \dots +\ln\left(1 + {1\over x_n}\right)$$

We know that $$x_k$$ is a natural number and therefore $$x_k > -1$$. Based on that we may use the following inequality:

$$\ln(1+x) < x, \; \forall x>-1$$

Thus:

$$\ln y_n = \sum_{k=1}^n\left(\ln\left(1+{1\over x_k}\right)\right) <\sum_{k=1}^n{1\over x_k} = S_n$$

But we know that $$S_n$$ is bounded, now given $$S_n < M$$: $$\ln y_n

Thus $$e^M$$ is some constant being an upper bound for $$y_n$$ and $$y_n > 0$$. So finally:

$$0

Question:

I would like to know whether my proof is valid and find(if possible) a precalculus way of proving this, since $$\ln(1+x) < x$$ requires derivatives to be proven.

• Your proof looks fine to me. Commented Nov 1, 2018 at 13:44
• There are probably a dozen proofs of $\ln(1+x) < x$, and what you consider “precalculus” might depend on your definition of the logarithm. Commented Nov 1, 2018 at 13:54
• @MartinR for my case calculus start with the definition of limits. Commented Nov 1, 2018 at 13:55
• @MartinR I use the following $a^{log_aN} = N$. So $a^x=N$ and $x=log_aN$. Are there other definition of $\log$ i do not know of? Commented Nov 1, 2018 at 14:00
• How do you define $e$ the natural base of logarithm without limit?
– user593746
Commented Nov 1, 2018 at 14:18

The OP's solution is fine, but it is technically not a precalculus solution. Here is a solution without even using logarithm. I only require that $$x_1,x_2,\ldots$$ are positive real numbers such that $$S_n$$ converges. (Well, I could also allow negative $$x_k$$ as long as $$\sum_{k=1}^\infty\frac{1}{x_k}$$ converges absolutely, but because the OP wants an elementary method, this more general assumption is not being made.)
Since $$S_n$$ is bounded, there can be only finitely many $$k$$ with $$0. We claim that, for some $$m$$, $$\sum_{k=m+1}^n\frac{1}{x_k}<\frac12$$ for every $$n>m$$. Suppose contrary that, for each $$m$$, there exists $$n_m>m$$ for which $$\sum_{k=m+1}^{n_m}\frac{1}{x_k}\geq \frac12.$$ Set $$t_0=0$$, $$t_1=n_{t_0}$$, $$t_2=n_{t_1}$$, $$t_3=n_{t_2}$$, $$\ldots$$. Then, $$\sum_{k=t_j+1}^{t_{j+1}}\frac{1}{x_k}\geq \frac12$$ for $$j=0,1,2,\ldots$$. In particular, $$S_{t_l}=\sum_{k=1}^{t_l}\frac{1}{x_k}=\sum_{j=0}^{l-1}\sum_{k=t_j+1}^{t_{j+1}}\frac{1}{x_k}\geq \sum_{j=0}^{l-1}\frac12\geq \frac{l}{2}$$ for all $$l=1,2,3,\ldots$$. This contradicts the assumption that $$S_n$$ is bounded.
So, there does exist $$m$$ such that $$\sum_{k=m+1}^n\frac{1}{x_k}<\frac12$$ for all $$n>m$$. Thus, $$y_n=\prod_{k=1}^m\left(1+\frac{1}{x_k}\right)\prod_{k=m+1}^n\left(1+\frac{1}{x_k}\right)$$ for $$n>m$$. Now, $$\prod_{k=m+1}^n\left(1+\frac{1}{x_k}\right)=\frac{\prod_{k=m+1}^n\left(1-\frac{1}{x_k^2}\right)}{\prod_{k=m+1}^n\left(1-\frac{1}{x_k}\right)}<\frac{1}{\prod_{k=m+1}^n\left(1-\frac{1}{x_k}\right)}.$$ By induction, $$\sum_{k=m+1}^n\left(1-\frac{1}{x_k}\right)\geq 1-\sum_{k=m+1}^n\frac{1}{x_k}>\frac12.$$ Hence, for $$n>m$$, $$y_n<2\prod_{k=1}^m\left(1+\frac{1}{x_k}\right)<\infty.$$ Because $$m$$ is fixed, the sequence $$\left(y_n\right)$$ is indeed bounded.
• Well, if $x_k$ can be negative but $\sum\limits_{k=1}^\infty\,\dfrac1{x_k}$ converges absolutely, then you can replace $x_k$ by $\left|x_k\right|$ in certain places to establish that $\left|y_n\right|$ is bounded. Commented Nov 1, 2018 at 15:49