Let $(x_n),(y_n)$ be two sequences of positive real numbers, with $x_n \to \infty$ and $$S_n=\frac{x_n}{x_n+y_1}+\frac{x_n}{2x_n+y_2}+\dots+\frac{x_n}{nx_n+y_n}$$ Prove that $\lim_{n\to \infty}S_n = \infty$

I tried to write the given sum as $S_n=x_n \left(\frac{1}{x_n+y_1}+\frac{1}{2\left(x_n+\frac{y_2}{2}\right)}+\dots+\frac{1}{n\left(x_n+\frac{y_n}{n}\right)} \right)$ and managed to prove the claim when $\left(\frac{y_n}{n}\right)$ is bounded. For the case when it is unbounded, I tried to use the sequence $a_n=\max \{\frac{y_1}{1}, \dots , \frac{y_n}{n} \}$, for which $a_n \to \infty$, in order to get $S_n \geq \frac{x_n}{x_n+a_n}\left(1+\frac{1}{2}+\dots+\frac{1}{n} \right)$, but I couldn't finish.

Also, I tried writing the sum as $S_n=\frac{1}{1+\frac{y_1}{x_n}}+\frac{1}{2+\frac{y_2}{x_n}}+\dots+\frac{1}{n+\frac{y_n}{x_n}}$, but nothing came out of it...

  • $\begingroup$ Did you just change $y_i$ to $y_n$? Because that changes everything about your initial problem and it's important to let others know it. $\endgroup$ – stressed out Nov 11 '17 at 13:45
  • $\begingroup$ I'm sorry, it never had any $i$-s in it, maybe you misread it. I removed the supposition I had made at the end that $\frac{y_n}{n}$ should be bounded. $\endgroup$ – Shroud Nov 11 '17 at 13:58

Let $i=10^{100000}$, choose $N$ such that for each $n\geq N$, we have $$\frac{y_1}{x_n}< 1,\quad \frac{y_2}{x_n}< 1, \quad \cdots \quad \frac{y_i}{x_n}<1$$

This is possible because $x_n\to\infty$. Then for $n\geq N$, we have $$\begin{aligned}S_n &> \frac{x_n}{x_n + y_1} + \frac{x_n}{2x_n + y_2}+\cdots+\frac{x_n}{ix_n + y_i} \\ &> \frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{i+1} = \frac{1}{2}+\frac{1}{3} + \cdots + \frac{1}{10^{100000}+1} \end{aligned}$$ Let $i$ be still larger number to conlcude $S_n\to \infty$.

  • $\begingroup$ Ah. This is cool. It's almost what I was doing on paper for solving the problem before you posted it. By the way, does the statement at the end of my argument hold as well? Do we have a theorem about the convergence behavior of two infinite series whose sequences are asymptotically equivalent? $\endgroup$ – stressed out Nov 11 '17 at 13:02
  • $\begingroup$ I think that the righthandside of the last inequality should be $\dfrac 12 + \dfrac 13 + \ldots + \dfrac{1}{i+1}$ $\endgroup$ – timon92 Nov 11 '17 at 13:04
  • $\begingroup$ @timon92 yeah, thanks for ponting this out. $\endgroup$ – pisco Nov 11 '17 at 13:07
  • 2
    $\begingroup$ I think it would be clearer to write: Fix any number $M$. Pick $i$ such that $\dfrac 12+\dfrac 13 + \ldots+\dfrac{1}{i+1} > M$. Choose $N$ such that [blah blah blah]. Therefore for $n>N$ we have $S_n>M$. Since we chose $M$ arbitrarily, $S_n \to \infty$. $\endgroup$ – timon92 Nov 11 '17 at 13:41
  • $\begingroup$ @pisco125 I think now, after your fixing, I see that it's indeed true. Nice! +1 $\endgroup$ – Michael Rozenberg Nov 11 '17 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.