Is $\lim_{n\to\infty}a\log n-\sum_{k=n^a+1}^{n^{a+1}}\frac{1}{k}=0$ for each integer $a\geq 1$? I believe that for each integer $a\geq 1$
$$\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log\left(n+\frac{1}{n^a}\right)\right),$$
where $\gamma$ is the Euler's constant, when I've use an idea that I've read in a paper. 
First, I compute 
$$\lim_{n\to\infty}\log\left(\frac{n}{n+\frac{1}{n^a}}\right)=0,$$
because 
$$\log\left(\lim_{n\to\infty}\frac{n^{a+1}}{n^{a+1}+1}\right)=\log 1,$$
where $a\geq 1$ is a fixed integer; secondly I write 
$$\sum_{k=1}^n\frac{1}{k}-\log\left(n+\frac{1}{n^a}\right)=\sum_{k=1}^{n^{a+1}}\frac{1}{k}-\log\left(n^{a+1}-1\right)+a\log n-\sum_{k=n^a+1}^{n^{a+1}}\frac{1}{k}.\tag{1}$$
Then when I take the limit as $n\to\infty$, previous formula finally shows that 
$$\lim_{n\to\infty}a\log n-\sum_{k=n^a+1}^{n^{a+1}}\frac{1}{k}=0.$$

Question. Are rights my computations? Can you give a different proof of 
  $$\lim_{n\to\infty}a\log n-\sum_{k=n^a+1}^{n^{a+1}}\frac{1}{k}=0,$$
  when $a\geq 1$ is a fixed integer? Thanks in advance.

 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\mbox{Note that}\quad \sum_{k\ =\ n^{a}\ +\ 1}^{n^{a + 1}}{1 \over k} & =
\sum_{k = 1}^{n^{a + 1}}{1 \over k} - \sum_{k = 1}^{n^{a}}{1 \over k} =
H_{n^{a + 1}}\ -\ H_{n^{a}}
\end{align}
where $H_{m}$ is a Harmonic Number which can be expressed in terms of the Digamma function $\Psi$. Namely, $H_{m} = \Psi\pars{m + 1} + \gamma$. $\gamma$ is the Euler-Mascheroni constant. Then,
\begin{align}
\sum_{k\ =\ n^{a}\ +\ 1}^{n^{a + 1}}{1 \over k} & =
\Psi\pars{n^{a + 1} + 1} - \Psi\pars{n^{a} + 1} = 
\Psi\pars{n^{a + 1}} + {1 \over n^{a + 1}} - \Psi\pars{n^{a}}
- {1 \over n^{a}} 
\end{align}
In this expression we used the Digamma Reccurrence Formula
$\ds{\Psi\pars{z + 1} = \Psi\pars{z} + {1 \over z}}$. The $\Psi$ asymptotic behaviour is given by:
$$
\Psi\pars{z} \sim \ln\pars{z} - {1 \over 2z} + \mathrm{O}\pars{z^{-2}}
$$ 
and
\begin{align}
\sum_{k\ =\ n^{a}\ +\ 1}^{n^{a + 1}}{1 \over k} & \sim
\ln\pars{n^{a + 1}} - \ln\pars{n^{a}} + \cdots = a\ln\pars{n} + \cdots
\end{align}
The 'remaining terms' $\ds{\cdots}$ go to $0$ when $n \to \infty$ such that
$$
\color{#f00}{\lim_{n \to \infty}\bracks{%
a\ln\pars{n} - \sum_{k\ =\ n^{a}\ +\ 1}^{n^{a + 1}}{1 \over k}}}
= \color{#f00}{0}
$$
A: I think there is a mistake. Using Abel's summation we have $$S=\sum_{k=n^{a}+1}^{n^{a+1}}\frac{1}{k}=\sum_{k=n^{a}+1}^{n^{a+1}}1\cdot\frac{1}{k}=\sum_{k=1}^{n^{a+1}}1\cdot\frac{1}{k}-\sum_{k=1}^{n^{a}}1\cdot\frac{1}{k}=\int_{n^{a}}^{n^{a+1}}\frac{\left\lfloor t\right\rfloor }{t^{2}}dt
 $$ where $\left\lfloor t\right\rfloor 
 $ is the floor function. Since $\left\lfloor t\right\rfloor =t+O\left(1\right)
 $ we have $$S=\log\left(n^{a+1}\right)-\log\left(n^{a}\right)+O\left(\frac{1}{n^{a}}\right)
 $$ $$=\log\left(n\right)+O\left(\frac{1}{n^{a}}\right)
 $$ so $$a\log\left(n\right)-S=\log\left(n^{a-1}\right)+O\left(\frac{1}{n^{a}}\right)\rightarrow\infty
 $$ if $a>1
 $. If $a=1
 $ we have $$\log\left(n\right)-S=O\left(\frac{1}{n}\right)
 $$ so the limit is $0$.
