Let $Y_{n} > 0$ for all $ n\in \mathbb{N} $, with $\sum{Y_{n}}= +\infty$.

If $\displaystyle\lim\limits_{n\rightarrow \infty}\frac{X_{n}}{Y_{n}}= a$ then $\displaystyle\lim\limits_{n\rightarrow\infty}\frac{X_{1}+X_{2}+X_{3}+\dots+X_{n}}{Y_{1}+Y_{2}+Y_{3}+\dots+Y_{n}}= a$

I need a idea for solution.

can i supose that $\lim\limits_{n\rightarrow\infty}Y_{n}$ exist ?

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    $\begingroup$ The following should probably work: fix $\epsilon > 0$, for $n > N $ we have $a - \epsilon < \frac{X_n}{Y_n} < a + \epsilon$, and since $\sum Y_n = \infty$ you can pretty much ignore the first $N$ terms, which puts the sum-fraction within $\epsilon$ of $a$. $\endgroup$ Jan 19, 2012 at 4:16
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    $\begingroup$ I suspect that nailing down the details will be somewhat complicated, but intuitively you have that $X_n\approx aY_n$ for sufficiently large $n$, and since $\sum Y_n=\infty$, any "initial terms" $X_1+\cdots+X_n$ and $Y_1+\cdots +Y_n$ will be negligible; so at the limit you will "essentially" just have $a$ times the sum of the $Y_i$s divided by the sum of the $Y_i$s. $\endgroup$ Jan 19, 2012 at 4:20
  • $\begingroup$ The epsilon-delta approach works fine without any extra hypothesis (like that $(Y_n)$ should converge), except that, as always with limits of sequences, it is rather an epsilon-N proof... $\endgroup$
    – Did
    Jan 19, 2012 at 6:44
  • $\begingroup$ This is basically Stolz-Cesaro theorem, although in a slightly different form as given at Wikipedia. It was mentioned in several questions, but I am not sure if we already have a proof in some of the answers at this site. $\endgroup$ Jan 19, 2012 at 7:00

2 Answers 2


$\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}$This observation is perhaps trivial, but I think it's useful to mention that this is just a different form of Stolz-Cesaro theorem.

Let us have a look at these two claims.

Claim 1: Let $(x_n)$ and $(y_n)$ be real sequences. Let $(y_n)$ be positive, strictly increasing and $\limti n y_n=+\infty$. If there exists a limit $$\limti n\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=L,$$ then $$\limti n\frac{x_n}{y_n}=L.$$

Claim 2: Let $(a_n)$ and $(b_n)$ be real sequences such that $b_n>0$ for each $n\in\mathbb N$ and $\sum b_n=+\infty$. If there exists a limit $$\limti n\frac{a_n}{b_n}=L,$$ then $$\limti n\frac{\sum_{k=1}^n a_k}{\sum_{k=1}^n b_k}=L.$$

By taking $a_n=x_{n+1}-x_n$ and $b_n=y_{n+1}-y_n$ we get Claim 2 from Claim 1. Conversely, by taking the partial sums $x_n=\sum_{k=1}^n a_n$ and $y_n=\sum_{k=1}^n b_n$ we get Claim 1 from Claim 2.

Basically with the same proof a slightly more general result can be shown. The proof can be found e.g. here.

Theorem Let $(x_n)$ and $(y_n)$ be real sequences. Let $(y_n)$ be positive, strictly increasing and $\limti n y_n=+\infty$. $$ \liminf\frac{x_{n+1}-x_n}{y_{n+1}-y_n}\le \liminf\frac{x_n}{y_n}\le \limsup\frac{x_n}{y_n}\le \limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}. $$


it sounds like you'd like a less lazy hint, so here goes:

Fix $\epsilon > 0$, we'll show that the partial sums $\frac{X_1 + \ldots + X_n}{Y_1 + \ldots +Y_n}$ are eventually trapped in $[a-\epsilon, a+ \epsilon]$, which tells us exactly what we want.

Right, for that same $\epsilon$, we know that for $n > N$ we have $$a - \epsilon < \frac{X_n}{Y_n} < a + \epsilon$$so our partial sums are bounded by: $$\frac{(X_1 + \ldots + X_N) + (a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n} < \frac{X_1 + \ldots + X_n}{Y_1 + \ldots +Y_n} < $$$$\frac{(X_1 + \ldots + X_N) + (a+\epsilon)Y_{N+1} + \ldots + (a + \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$(I just literally stuck in our estimate) Now, the idea is that the left and right hand bounds converge to $a - \epsilon$, $a + \epsilon$ respectively.

I'll just show one of them, $a - \epsilon$. First ``tear off" the first $N$ terms, since $\sum Y_n = \infty$ they'll dissappear in the limit, that is we have: $$\lim_n \frac{(X_1 + \ldots + X_N) + (a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n} = $$$$\lim_n \frac{X_1 + \ldots + X_N}{Y_1 + \ldots + Y_n} + \lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$$$0 + \lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$Right, now to manage this guy, notice that if you forgot about $Y_1 + \ldots + Y_N$ in the denominator, you'd just have $a-\epsilon$, which is what we want, and sure enough: $$\lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n} = \lim_n \frac{(a - \epsilon)Y_{N+1} + \ldots + (a-\epsilon)Y_n}{Y_{N+1} + \ldots + Y_n} \frac{Y_{N+1} + \ldots + Y_n}{Y_1 + \ldots + Y_n}$$$$ = (a-\epsilon) \lim_n \frac{Y_{N+1} + \ldots + Y_n}{Y_1 + \ldots + Y_n} = (a - \epsilon) \cdot 1$$which is what we wanted.

Comment or something if you have any questions etc. I hope this helps mate :)

  • $\begingroup$ ok, i thought go at limit without suppose the existence of itself. thank you for you answer, helped me a lot. I will do the right hand of the inequality. only question: why supose the existence the limit of $Y_{n}$, be because it diverges? again thanks. $\endgroup$ Jan 19, 2012 at 7:20
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    $\begingroup$ "i thought go at limit without suppose the existence of itself" This argument doesn't assume the limit $\frac{X_1 + \ldots + X_n}{Y_1 + \ldots + Y_n}$ exists, wherever you see $$\lim_n$$just add "if it exists". As for "why supose the existence the limit $Y_n$", this argument doesn't use the convergence of the $Y_n$ anywhere. So you don't need to suppose $\lim_n Y_n$ exists. $\endgroup$ Jan 19, 2012 at 7:43
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    $\begingroup$ @uncookedfalcon : Why $\lim_n \frac{Y_{N+1}+\dots +Y_n}{Y_1+\dots +Y_n}=1$? Sorry for the question... $\endgroup$
    – Derso
    Jan 22, 2015 at 0:03

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