Problem 39

ThePerfectHacker · #1 $Old$ October 29th, 2007, 09:20 PM

Let $a,b\in \mathbb{Z}^+$ . For each positive integer $n$ let $H_n(a,b) = \frac{1}{a+b}+\frac{1}{2a+b}+...+\frac{1}{na+b}$ .
Find the limit,
$\lim \ \frac{H_n(a,b)}{H_n(c,d)}$

(Where $c,d$ are possibly different integers defining a different sequence).

kalagota · #2 $Old$ October 31st, 2007, 02:20 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

Let $a,b\in \mathbb{Z}^+$ . For each positive integer $n$ let $H_n(a,b) = \frac{1}{a+b}+\frac{1}{2a+b}+...+\frac{1}{na+b}$ .
Find the limit,
$\lim \ \frac{H_n(a,b)}{H_n(c,d)}$

(Where $c,d$ are possibly different integers defining a different sequence).

is this lim as n approaches infinity?

ThePerfectHacker · #3 $Old$ October 31st, 2007, 07:45 AM

Quote:

Originally Posted by kalagota $View Post$

is this lim as n approaches infinity?

When you are dealing with sequences that is the only type of limit you can have. (What else can you possibly approach

)

galactus · #4 $Old$ October 31st, 2007, 10:26 AM

Maybe I'm off base here, PH,:

$\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{na+b}\right)=\frac{1}{a}$

$\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{nc+d}\right)=\frac{1}{c}$

So, we have: $\frac{\frac{1}{a}}{\frac{1}{c}}=\boxed{\frac{c}{a}}$

ThePerfectHacker · #5 $Old$ October 31st, 2007, 11:09 AM

Quote:

Originally Posted by galactus $View Post$

Maybe I'm off base here, PH,:

$\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{na+b}\right)=\frac{1}{a}$

$\lim_{n\rightarrow{\infty}}\sum_{k=1}^{n}\left(\frac{1}{nc+d}\right)=\frac{1}{c}$

The sums diverges. Because,
$0\leq \frac{1}{n+n}\leq \frac{1}{na+b}$ for sufficiently large $n$ .

And $\sum_{n=1}^{\infty}\frac{1}{2n}$ does not converge, it is the harmonic series.

red_dog · #6 $Old$ November 1st, 2007, 07:41 AM

Let $x_n=H_n(a,b), \ y_n=H_n(c,d)$ .
$(y_n)$ is ascending and unbounded.
By Stolz-Cesaro, we have
$\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{\frac{1}{a(n+1)+b}}{\frac{1}{c(n+1)+d}}=\lim_{n\to\infty}\frac{cn+c+d}{an+a+b}=\frac{c}{a}$ .

Then $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\frac{c}{a}$

ThePerfectHacker · #7 $Old$ November 3rd, 2007, 10:07 PM

Quote:

Originally Posted by red_dog $View Post$

Let $x_n=H_n(a,b), \ y_n=H_n(c,d)$ .
$(y_n)$ is ascending and unbounded.
By Stolz-Cesaro, we have
$\displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{\frac{1}{a(n+1)+b}}{\frac{1}{c(n+1)+d}}=\lim_{n\to\infty}\frac{cn+c+d}{an+a+b}=\frac{c}{a}$ .

Then $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\frac{c}{a}$

I can do it without Stolz-Cesaro. Want to try?

ThePerfectHacker · #8 $Old$ November 10th, 2007, 08:49 PM

The key step is to note that $H_n(a,b)\sim \ln (an+b)$ . That means, $\lim \ \frac{H_n(a,b)}{H_n(c,d)} = \lim \ \frac{\ln (an+b)}{\ln (cn+d)}=\frac{c}{a}$ .