# Prove that $\lim_{n\to \infty}S_n = \infty$

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...

• 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. – stressed out Nov 11 '17 at 13:45
• 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. – 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 $S_n\to \infty$.
• I think that the righthandside of the last inequality should be $\dfrac 12 + \dfrac 13 + \ldots + \dfrac{1}{i+1}$ – timon92 Nov 11 '17 at 13:04
• 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, $S_n \to \infty$. – timon92 Nov 11 '17 at 13:41