1. ## Problem 39

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).

2. Originally Posted by ThePerfectHacker
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?

3. Originally Posted by kalagota
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 )

4. Maybe I'm off base here, PH,:

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

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

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

5. Originally Posted by galactus
Maybe I'm off base here, PH,:

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

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

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

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

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

Then $\displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=\fra c{c}{a}$
I can do it without Stolz-Cesaro. Want to try?

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