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What is the limit of $H_n - H_{n/2}$ for $n\to\infty$? $H_n$ is the nth Harmonic number.

And in general, what is the limit of $H_n - H_{a*n}$ for $n\to\infty$? $0 < a < 1$.

Summing these with Python gives the following results.

..............a=1/2...........a=1/3...........a=1/4...........a=1/5 n=1,000.......0.69364743056...1.10061495867...1.38779561112...1.61143991243 n=10,000......0.69319718306...1.09881231534...1.38644437362...1.60963793243 n=100,000.....0.693152180585..1.09863228893...1.38630936124...1.60945791263 n=1,000,000...0.69314768056...1.09861428867...1.38629586112...1.60943991244 n=10,000,000..0.69314723056...1.09861248867...1.38629451112...1.60943811243

These sums are getting smaller, and they can't go to zero, so they have limits. What would be their exact mathematical expressions?

Thank you.

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For example, if n=1,000, a=1/2, then 1/500 + 1/501 + ... + 1/1,000. It would be H_1000 - H_499, where H_n is the nth Harmonic number. –  Phillip Jan 25 '12 at 2:38
    
I'm not sure I understand your formula, but those sequences of numbers clearly converge to $\ln(2)$, $\ln(3)$, $\ln(4)$, $\ln(5)$, etc. –  Byron Schmuland Jan 25 '12 at 2:42
    
Thank you very much! That answers my question. –  Phillip Jan 25 '12 at 2:52
    
On page 82 of the January Monthly, try to find the limit as $n \rightarrow \infty$ of $$ 2 H_n - H_{n^2 + n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \frac{1}{n + 1 } - \cdots - \frac{1}{n + n^2 } \; ? $$ Evidently submitted by Juozas Juvencijus Macys. I changed it slightly. –  Will Jagy Jan 25 '12 at 6:43
    
Is it not just $\gamma$ ? –  Ragib Zaman Jan 25 '12 at 7:23

2 Answers 2

up vote 4 down vote accepted

For easiness, say $b=a^{-1}$. Then, the limit the OP is interested in is $ \lim{}_{n\rightarrow \infty }\sum_{k=n/b}^n \frac{1}{k}$ (assuming $b | n$ of course). We can change the sum to $$\sum_{k=0}^{n-n/b}\frac{b}{bk+n}=\sum_{k=0}^{n-n/b}\frac{1}{\frac{bk}{n}+1}\cdot\frac{b}{n}$$

This is a riemann sum where $dx = b/n$, and by taking limits we see that $$\lim_{n\rightarrow \infty}\sum_{k=0}^{n-n/b}\frac{1}{\frac{bk}{n}+1}\cdot\frac{b}{n} = \int_0^{b-1} \frac{1}{x+1}dx = \log(b)$$

there are a few math GRE subj problems like this ;)

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Thank you very much. –  Phillip Jan 25 '12 at 3:59

Worth remembering is that $H_n=\log(n)+\gamma+o(1)$ where $\gamma=0.577\ldots$ is Euler-Mascheroni constant. Hence, for every $a\gt0$ and $b\gt0$, $$ H_{\lfloor an\rfloor}-H_{\lfloor bn\rfloor}=\log(an)+\gamma-\log(bn)-\gamma+o(1)=\log(a)-\log(b)+o(1), $$ which means exactly that $H_{\lfloor an\rfloor}-H_{\lfloor bn\rfloor}\to\log(a/b)$. In your case, $H_n-H_{\lfloor n/2\rfloor}\to\log(2)$.

A proof of the existence of $\gamma$ uses the expression $$H_n-\log(n)=\sum\limits_{k=1}^{n}u_k\quad\text{with}\quad u_k=\frac1k-\log\left(1+\frac1k\right). $$ Then the series expansion of the logarithm shows that $x-\frac12x^2\lt\log(1+x)\lt x$ for every $x$ in $(0,1)$, hence $0\lt u_k\lt\frac1{2k^2}$ for every $k\geqslant1$ and the series $\sum\limits_{k}u_k$ converges.

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Thanks a lot, I learnt something here. –  Phillip Jan 28 '12 at 2:56
    
You are welcome. Unrelated: next time, you might wish to wait more than 97 minutes before accepting an answer. –  Did Jan 28 '12 at 8:30

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