Is $\lim_{n\to\infty}a\log n-\sum_{k=n^a+1}^{n^{a+1}}\frac{1}{k}=0$ for each integer $a\geq 1$?

Question

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.

Answer 1

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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

$$ \color{#f00}{\lim_{n \to \infty}\bracks{% a\ln\pars{n} - \sum_{k\ =\ n^{a}\ +\ 1}^{n^{a + 1}}{1 \over k}}} = \color{#f00}{0} $$

Answer 2

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 $0$.