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$\require{cancel}$I need to evaluate the following limit: $$\lim_{n\to\infty}\sum_{np}^{nq}\frac1k \quad \text{where $\quad p\geq1,q\geq1$}$$

The way I tackled this problem is as follows:

Recall the relation:

$$\sum_{k=1}^n\frac1k = \ln(n)+\gamma+o(1) \quad \text{as $n\to\infty$}$$

Therefore, assuming $q\geq p$, the initial limit can be expressed as:

$$\begin{align*} \lim_{n\to\infty}\sum_{np}^{nq}\frac1k &= \lim_{n\to\infty}\left(\sum_{k=1}^{nq}\frac1k-\sum_{k=1}^{np-1}\frac1k\right)\\ &=\lim_{n\to\infty}\left(\ln(nq)+\cancel{\gamma}-\ln(np-1)\cancel{-\gamma}+o(1)\right)\\ &= \lim_{n\to\infty}\ln\left(\frac{nq}{np-1}+o(1)\right)=\ln\left(\lim_{n\to\infty}\left(\frac{\cancel{n}q}{\cancel{n}\left(p-\frac1n\right)}+o(1)\right)\right)=\boxed{\ln\left(\frac{q}{p}\right)} \end{align*}$$

Are the result and procedure correct? Also, do I need to prove the initial identity (for a Homework Assignment)?

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    $\begingroup$ Note that cancelling the $o(1)$ terms doesn't make sense (for instance, $1/n$ and $1/2n$ are both $o(1)$, but $1/n - 1/2n \neq 0$). What you can say, however, is $o(1) -o(1) = o(1)$. This still goes away in the limit, so your answer remains fine. $\endgroup$ – stochasticboy321 Mar 20 '17 at 2:46
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    $\begingroup$ @FelixMarin Why the $+1$ in the logarithm? Take a look at: Euler-Mascheroni Constant. It does not include the $+1$ term when defining the constant $\endgroup$ – DMH16 Mar 20 '17 at 2:46
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    $\begingroup$ It doesn't matter because $\lim_{n \to \infty}\left[\ln\left(n + 1\right) - \ln\left(n\right)\right] = 0$. I just was using the relation $H_{n} = \Psi\left(n + 1\right) + \gamma$ (see this link) and $\Psi\left(m\right) \sim \ln\left(m\right)$ as $m \to \infty$. In your proof, which is CORRECT, it's not important. $\endgroup$ – Felix Marin Mar 20 '17 at 2:54
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    $\begingroup$ Whether or not you need to prove the initial equation depends on the course, the instructor, and what results you are allowed to "import". Let $F(n)=-\ln n+\sum_{j=1}^n1/j$. Then $| F(n+1)-F(n)|=$ $|\ln (1-1/(n+1))+1/(n+1)|=$ $(n+1)^{-2}(1/2+1/3(n+1)+1/4(n+1)^2+...)$ $< (n+1)^{-2}$ so $F(n)=\gamma +O(n^{-1})$ as $n\to \infty.$ $\endgroup$ – DanielWainfleet Mar 20 '17 at 3:22
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    $\begingroup$ If you substitute the equation for the relation into each of the summations in the 1st line of your proof, the term o(1) will appear twice. Of course o(1)+o(1)=o(1), so the two terms will telescope into one term in the 3rd line. Whether this needs to be written in detail that way depends on your audience, as only a newcomer to that notation would need to see it. But whether your instructor wants you to show that YOU do undertand it, I cannot say. However I would write it in. You cannot be faulted for too much rigor, except on aesthetic grounds. $\endgroup$ – DanielWainfleet Mar 20 '17 at 3:37
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The approach in the OP is fine.

I thought it might be instructive to present another way forward that doesn't rely on the asymptotic expansion of the Harmonic number, but rather relies on only straightforward analysis and Riemann sums. To the end, we proceed.

We simply write

$$\begin{align} \sum_{k=np}^{nq}\frac{1}{k}&=\sum_{k=0}^{n(q-p)}\frac{1}{k+np}\\\\ &=\frac{1}{np}\sum_{k=0}^{n(q-p)}\frac{1}{1+\frac k{np}}\tag 1 \end{align}$$

We recognize that the right-hand side of $(1)$ is the Riemann sum for

$$\begin{align} \frac1p\int_0^{q-p}\frac{1}{1+x/p}\,dx&=\log\left(1+\frac{q-p}{p}\right)\\\\ &=\log(q/p) \end{align}$$

Hence, we see that the coveted limit is

$$\lim_{n\to \infty}\sum_{k=np}^{nq}\frac{1}{k}=\log(q/p)$$

as expected!

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Going directly from $\ln(x) =\int_1^x \frac{dt}{t} $, we have $\ln(k+1)-\ln(k) =\int_k^{k+1} \frac{dt}{t} $, so $\frac1{k+1} < \ln(k+1)-\ln(k) < \frac1{k} $.

Summing, $$\sum_{k=np}^{nq-1}\frac1{k+1} <\sum_{k=np}^{nq-1} (\ln(k+1)-\ln(k)) <\sum_{k=np}^{nq-1} \frac1{k} $$ so that $$\sum_{k=np+1}^{nq}\frac1{k} =\sum_{k=np}^{nq}\frac1{k}-\frac1{np} < \ln(nq)-\ln(np) =\ln(q/p) <\sum_{k=np}^{nq-1} \frac1{k} =\sum_{k=np}^{nq} \frac1{k}-\frac1{nq} .$$

Therefore $$-\frac1{np} < \ln(q/p)-\sum_{k=np}^{nq}\frac1{k} < -\frac1{nq} $$ or $$\frac1{nq} < \sum_{k=np}^{nq}\frac1{k}-\ln(q/p) < \frac1{np} .$$

Now let $n \to \infty$.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\lim_{n \to \infty}\sum_{np}^{nq}{1 \over k}:\ {\large ?} \quad\mbox{where}\quad p \geq 1\,,\ q \geq 1}$.

\begin{align} \lim_{n \to \infty}\sum_{np}^{nq}{1 \over k} & = \lim_{n \to \infty}\pars{\sum_{k = 1}^{nq}{1 \over k} - \sum_{k = 1}^{np - 1}{1 \over k}} = \lim_{n \to \infty}\pars{H_{nq} - H_{np - 1}}\qquad \pars{~H_{m}:\ Harmonic\ Number~} \\[5mm] & = \lim_{n \to \infty}\bracks{\Psi\pars{nq + 1} - \Psi\pars{np}}\qquad \pars{~\Psi:\ Digamma\ Function~} \\[5mm] & \lim_{n \to \infty}\bracks{\ln\pars{nq + 1} - \ln\pars{np}} = \bbx{\ds{\ln\pars{q \over p}}} \end{align}

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