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:

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


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

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

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

$$ 0<y_n < e^M $$


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.

  • 2
    $\begingroup$ Your proof looks fine to me. $\endgroup$ Commented Nov 1, 2018 at 13:44
  • $\begingroup$ There are probably a dozen proofs of $\ln(1+x) < x$, and what you consider “precalculus” might depend on your definition of the logarithm. $\endgroup$
    – Martin R
    Commented Nov 1, 2018 at 13:54
  • $\begingroup$ @MartinR for my case calculus start with the definition of limits. $\endgroup$
    – roman
    Commented Nov 1, 2018 at 13:55
  • $\begingroup$ @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? $\endgroup$
    – roman
    Commented Nov 1, 2018 at 14:00
  • 2
    $\begingroup$ How do you define $e$ the natural base of logarithm without limit? $\endgroup$
    – user593746
    Commented Nov 1, 2018 at 14:18

1 Answer 1


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 $0<x_k<1$. 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.

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Nov 1, 2018 at 15:49
  • 1
    $\begingroup$ I think you made a typo. I fixed it, and hopefully, you wouldn't mind. $\endgroup$ Commented Nov 1, 2018 at 16:42

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