10

I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.

My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$.

Is that ok?

Thanks.

Martin Sleziak
  • 53,687
user1212
  • 101

5 Answers5

9

By the Stolz-Cesaro theorem from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem, $$\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{2n+3}}=2.$$

Martin Sleziak
  • 53,687
xpaul
  • 44,000
6

Hint

The numerator is $H_n$ and the denominator is $H_{2n+1}-\frac12H_n$.

Also,

$$\frac{H_n}{H_{2n+1}-\frac12H_n}=\frac1{-\frac12+\frac{H_{2n+1}}{H_n}}$$

and $$H_n\sim\ln n$$

ajotatxe
  • 65,084
5

If the numerator is always twice the denominator, then this works provided you include a proof of that. Let's try it when $n=2$: $$ \frac{1+\frac12}{1+\frac13+\frac15} = \frac{30+15}{30+10+6}= \frac{45}{46}\ne 2. $$ Let's try it when $n=3$: $$ \frac{1+\frac12+\frac13}{1+\frac 13+\frac15+\frac17}= \frac{210+105+70}{210+70+42+30} = \frac{385}{352} \ne 2. $$

Michael Hardy
  • 1
  • but my attempt is for on infinity. – user1212 Jun 02 '15 at 18:56
  • 3
    Your attempt appears to replace $\displaystyle\frac{1 +\frac12+\frac13+\cdots+\frac1n}{1+\frac13 +\frac15+\cdots+\frac{1}{2n+1}}$ with $\dfrac{2s}s$ and THEN to take the limit as $n\to\infty$. When you change the given fraction to $(2s)/2$ BEFORE taking the limit as $n\to\infty$, you're saying $(2s)/s$ works for every finite value of $n$. But it doesn't. ${}\qquad{}$ – Michael Hardy Jun 02 '15 at 18:59
  • as $s_1=1=1/3+1/5+...$ and $s_2=1+1/2+1/3+1/4+...$ so $2s_2-2s_1=1+1/2+1/3+1/4+...$ so $s_2=2s_1$ – user1212 Jun 02 '15 at 19:00
  • that's not true because i got those sums by taking $n\to\infty$ – user1212 Jun 02 '15 at 19:02
  • The limit is outside the ratio, where the quantity $s$ (which, incidentally, you never defined explicitly) appears in both the numerator and denominator. What you write is what @MichaelHardy pointed out: you replace a finite sum by something, then take the limit. But the thing you had and what you replaced it with are not equal, so the proof is not valid. – Clement C. Jun 02 '15 at 19:04
  • 3
    If you write $s_1=1+\frac13+\frac15+\cdots$, that means $\displaystyle s_1=\lim_{n\to\infty}\left(1+\frac13+\frac15+\cdots+\frac1n\right)$. One problem with this is that that sum is $\infty$ rather than any finite number. Another is that you're working on $\displaystyle\frac{\lim_{n\to\infty}(1+\frac12+\frac13+\cdots+\frac1n)}{\lim_{n\to\infty}(1+\frac13+\frac15+\cdots+\frac1{2n+1})}$ rather than $\displaystyle\lim_{n\to\infty}\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$. The latter limit exists and is a finite number. The former is $\infty/\infty.\quad{}$ – Michael Hardy Jun 02 '15 at 19:06
  • i'm confused because i guess i can solve it by my method but i don't know how to do it correctly by your remarks and the other solutions are hard for me to understand. – user1212 Jun 02 '15 at 19:31
  • @user1212: the issue with your method is that it is not a valid proof: the "sum" $s$ you are using is infinite, and you cannot apply mathematical operations on it as you do. – Clement C. Jun 02 '15 at 19:33
  • @user1212 : Something you should know about limits is that if $f\to\infty$ and $g\to\infty$, then in some cases $\lim\dfrac f g$ exists and is a finite number, but $\dfrac{\lim f}{\lim g}$ just gives $\dfrac\infty\infty$, so that doesn't work as a way of finding the limit. The same sort of thing often happens when $f\to0$ and $g\to0$. ${}\qquad{}$ – Michael Hardy Jun 02 '15 at 19:44
  • now i got it, thank you for your Patience. – user1212 Jun 02 '15 at 19:55
2

I don't know if there's still any interest in this question from more than a year ago, but here's an elementary solution:

The numerator in the problem is $H_n=1+\frac12+\frac13+\dots+\frac{1}{n},$ the $n^{\text{th}}$ harmonic number.

The denominator in the problem is

\begin{align} 1+\frac13+\frac15+\dots+\frac1{2n+1}&=H_{2n+1}-\big(\frac12+\frac14+\frac16+\dots+\frac{1}{2n}\big) \\&=H_{2n+1}-\frac12 \big(1+\frac12+\frac13+\dots+\frac{1}{n}\big) \\&=H_{2n+1}-\frac12 H_n. \end{align}

Next note that \begin{align} H_{2n+1}-H_n&=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{2n+1} \\&\le\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{n+1} \scriptsize{\quad(n+1\text{ terms})} \\&=1. \end{align}

Of course, $H_{2n+1}-H_n$ is always positive, being a sum of one or more positive fractions, so we have

$$0\lt H_{2n+1}-H_n \le 1$$

for all non-negative integers $n.$

Since $H_n\to\infty$ as $n\to\infty,$ it follows that

$$ \lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}=0. $$

We can now solve the original problem:

\begin{align} \displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}&=\lim_{n\to\infty}\frac{H_n}{H_{2n+1}-\frac12 H_n} \\&=\lim_{n\to\infty}\frac{H_n}{\frac12 H_n+(H_{2n+1}-H_n)} \\&=\lim_{n\to\infty}\frac{1}{\frac12+\frac{H_{2n+1}-H_n}{H_n}} \\&=\frac{1}{\frac12+\lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}} \\&=\frac{1}{\frac12+0} \\&=2. \end{align}

Mitchell Spector
  • 9,917
  • 3
  • 16
  • 34
0

With the Euler-MacLaurin formula, we have

numerator = $\ln(n)+O(1)$ and denominator = $\frac12 \ln(2n+1) + O(1)$, so the limit is

$\lim_{n \to \infty}\frac{\ln(n)+O(1)}{\frac12 \ln(2n+1) + O(1)}= 2.$

Claudeh5
  • 580