$\lim_{n\to\infty} \sum_{r=1}^n \frac{n}{n+r} - n\log 2$

Question

Consider the following two sequences of reals $$a_n=\frac{1}{n+1}+\cdots +\frac{1}{2n}$$ and $$b_n=\frac{1}{n}$$ How would one prove that $$\lim_{n\to\infty }\frac {a_n-\log 2}{b_n}$$ is finite and non-zero? Actually this is part of one of my old questions: Rate of convergent of real Sequence. But I am unable to to prove that $$\lim_{n\to\infty} \frac{a_n - \log 2}{b_n} = \lim_{n\to\infty} \sum_{r=1}^n \frac{n}{n+r} - n\log 2 = -\frac{1}{4} \ne 0, \pm\infty$$ I tried it but didn't get the limit. Thank you .

Answer 1

Use the Stolz-Cesàro theorem, where you can show that $g_n = a_n-\log 2$ (also: $a_n$ is of the form $H_{2n}-H_n$) is such that: $$g_{n+1} - g_n = \dfrac{1}{2(n+1)(2n+1)}.$$ Likewise, $$b_{n+1} - b_n = -\dfrac{1}{n(n+1)}.$$

By the theorem, the required ratio is then of the form $$\lim_{n\rightarrow\infty}-\dfrac{n+1}{4n+2},$$ which evaluates to $-\dfrac{1}{4}$.

Answer 2

You can easily obtain more than the limit itself $$S_n= \sum_{r=1}^n \frac{n}{n+r} - n\log( 2)=n\Big(\sum_{r=1}^n \frac{1}{n+r} - \log( 2)\Big)$$ Using harmonic numbers $$\sum_{r=1}^n \frac{1}{n+r}=\psi (2 n+1)-\psi (n+1)=H_{2 n}-H_n$$ Using the asymptotics of harmonic numbers $$H_p=\log (p)+\gamma +\frac{1}{2p}-\frac{1}{12 p^2}+\frac{1}{120 p^4}+O\left(\frac{1}{p^5}\right)$$ Applying it twice $$S_n=-\frac{1}{4}+\frac{1}{16n}-\frac{1}{128 n^3} +O\left(\frac{1}{n^5}\right)$$ wich gives the limit, shows how it is approached and provides tight bounds.

Answer 3

Your $a_n$ is a Riemann sum for $f(x) =1/(1+x)$ over interval $[0,1]$ and you can notice that by rewriting it as $$a_n=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+(k/n)}$$ Next note that $\int_0^1 f(x) \, dx=\log 2$ and we have a general theorem which says that $$\lim_{n\to\infty} n\left(\frac{1}{n}\sum_{k=1}^nf(k/n)-\int_0^1 f(x) \, dx\right) =\frac{f(1)-f(0)}{2}$$ and you get your desired limit as $-1/4$.

The theorem mentioned above and its generalisation are discussed in this answer.