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\begin{center}
\vskip 1cm{\LARGE\bf On the Lcm-Sum Function}
\vskip 1cm
\large
Soichi Ikeda and Kaneaki Matsuoka\\
Graduate School of Mathematics\\
Nagoya University\\
Furocho Chikusaku Nagoya 464-8602\\
Japan\\
\href{mailto:m10004u@math.nagoya-u.ac.jp}{\tt m10004u@math.nagoya-u.ac.jp}\\
\href{mailto:m10041v@math.nagoya-u.ac.jp}{\tt m10041v@math.nagoya-u.ac.jp}\\
\end{center}

\vskip .2 in

\begin{abstract}
We consider a generalization of the lcm-sum function, and we
give two kinds of asymptotic formulas for the sum of that
function. Our results include a generalization of
Bordell\`{e}s's results and a refinement of the error estimate
of Alladi's result. We prove these results by the method similar
to those of Bordell\`{e}s.
\end{abstract}

\section{Introduction}
Pillai \cite{pillai} first defined the gcd-sum function
\[ g(n) = \sum_{j=1}^n \gcd(j,n) \]
and studied this function. The function $g(n)$ was
defined again by Broughan \cite{broughan1}.
Broughan considered
\[ G_{\alpha}(x) = \sum_{n \le x} n^{- \alpha} g(n) \]
for $\alpha \in \mathbb{R}$ and $x \ge 1$, and obtained
some asymptotic formulas for $G_{\alpha}(x)$. The function
$G_{\alpha}(x)$ was studied by some authors (see, for example,
\cite{broughan2, tanigawa}). Some generalizations of
the function $g(n)$ was considered (see, for example,
\cite{bordelles, toth}).

On the other hand, the lcm-sum function
\[ l(n) := \sum_{j=1}^n \lcm(j,n) \]
was considered by some authors. Alladi \cite{alladi} studied the sum
\[ \sum_{j=1}^n (\lcm(j,n))^r \qquad (r \in \mathbb{R}, \, r \ge 1) \]
and obtained
\begin{equation} \label{res_alladi}
  \sum_{n \le x} \sum_{j=1}^n (\lcm(j,n))^r =
     \frac{\zeta(r+2)}{2(r+1)^2 \zeta(2)} x^{2r+2} +
     O(x^{2r+1+ \epsilon}).
\end{equation}

We define the functions
\begin{gather}
  L_a(n) := \sum_{j=1}^n (\lcm(j,n))^a \notag \\
  T_a(x) := \sum_{n \le x} L_a(n) \notag
\end{gather}
for $a \in \mathbb{Z}$ and $x \ge 1$.

Bordell\`{e}s studied the sums $T_1(x)$ and $T_{-1}(x)$
and obtained
\begin{gather}
  l(n) = \frac{1}{2}((\mathrm{Id}^2 \cdot (\varphi + \tau_0))
    \ast \mathrm{Id})(n), \notag \\
  \sum_{n \le x} \sum_{j=1}^n \lcm(j,n) = \frac{\zeta(3)}{8 \zeta(2)} x^4
    + O(x^3 (\log x)^{2/3} (\log \log x)^{4/3}) \quad (x &gt; e),  \notag \\
  \sum_{n \le x} \sum_{j=1}^n \frac{1}{\lcm(j,n)} =
    \frac{(\log x)^3}{6 \zeta(2)} + \frac{(\log x)^2}{2 \zeta(2)}
    \biggl( \gamma + \log \biggl( \frac{\mathcal{A}^{12}}{2 \pi} \biggr)
    \biggr) + O( \log x), \notag
\end{gather}
where $\mathrm{Id}^a(n) = n^a$
($a \in \mathbb{Z}$),
\[ \tau_0(n) = \begin{cases}
                1, &amp; \text{if $n = 1$;} \\
                0, &amp; \text{otherwise},
              \end{cases} \]
$F*G$ is the usual Dirichlet convolution product, and
$\mathcal{A}$ is the Glaisher-Kinkelin constant
\cite[p.\ 25]{srivastava}). Gould and Shonhiwa \cite{gould}
stated that the $\log$-factors in the error term in the
second formula can be removed.

In this paper we study $T_a(x)$ for $a \ge 2$ and $a \le -2$.
The following theorems are our main results.
These results are proved by the methods similar to
those of Bordell\`{e}s \cite[Section 6]{bordelles}.

We write $f(x) = O(g(x))$, or equivalently $f(x) \ll g(x)$,
where there is a constant $C &gt; 0$ such that $|f(x)| \le C g(x)$
for all values of $x$ under consideration.

\begin{theorem} \label{th_conv}
Let $B_n$ be Bernoulli numbers defined by
\[ \frac{z}{e^z - 1} = \sum_{n=0}^{\infty} B_n \frac{z^n}{n!}. \]
If we define
\[ \varphi_k(n) := \sum_{d \mid n} \mu(d) \Bigl( \frac{d}{n} \Bigr)^k \]
and
\[ M_a(n) := \biggl( \mathrm{Id}^{2a} \cdot \biggl(
     \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1}
     \sum_{k=1}^{a-1} \binom{a+1}{k+1}
     B_{k+1} \varphi_k \biggr) \biggr)(n), \]
then for $a \in \mathbb{Z}$ we have
\[ L_a(n) = (M_a \ast \mathrm{Id}^a)(n). \]
\end{theorem}

\begin{theorem} \label{th_pos}
Let $x &gt; e$ and $a \in \mathbb{N}$. Then we have
\[ \sum_{n \le x} L_a(n) = \frac{\zeta(a+2)}{2(a+1)^2 \zeta(2)} x^{2a+2}
     + O(x^{2a+1} (\log x)^{2/3} (\log \log x)^{4/3})
     \quad (\text{as $x \to \infty$}), \]
where the implied constant depends on $a$.
\end{theorem}

\begin{theorem}  \label{th_neg}
Let $x \ge 1$ and $k \in \mathbb{N}$ with $k \ge 2$. Then we have
\begin{equation} \label{series_lkn}
 \sum_{n=1}^{\infty}L_{-k}(n)=\frac{\zeta(k)}{2} \biggl(
     1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr)
\end{equation}
and
\[ \sum_{n \le x}L_{-k}(n)=\frac{\zeta(k)}{2} \biggl(
     1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr) - 
     \frac{\zeta(k)x^{-k+1}\log x}{(k-1)\zeta(k+1)}+O(x^{-k+1})
     \quad (\text{as $x \to \infty$}), \]
where the implied constant depends on $k$.
\end{theorem}

We note that the function $L_a(n)$ is not multiplicative for
all $a \in \mathbb{Z} \setminus \{0\}$, but we can write
$L_a(n)$ ($a \ge 1$) explicitly by Dirichlet convolution.
In the proof of Theorem \ref{th_pos} we use this fact.
The error estimates in Theorem \ref{th_pos} are better than
(\ref{res_alladi}). Since we have
\[ g_r(n) := \sum_{j=1}^n (\gcd(j,n))^r &gt; \varphi(n) \]
for all $r \in \mathbb{R}$, the sum
\begin{equation*}  \label{sum_gcd_r}
  \sum_{n = 1}^{\infty} g_r(n)
\end{equation*}
is divergent for all $r$. Therefore the behavior of the
sum $T_a(x)$ ($a \in \mathbb{Z}$ and $a \le -2$) is
completely different from that of the sum
\[ \sum_{n \le x} g_a(n). \]

\section{Lemmas for the proof of theorems}
In this section, we collect some auxiliary results and definitions.

Let $B_n(x)$ be Bernoulli polynomials defined by
\[ \frac{z e^{xz}}{e^z - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{z^n}{n!}. \]
The following relations are well-known \cite[p.\ 59]{srivastava}.
\begin{gather}
  B_n(x + 1) - B_n(x) = n x^{n-1},  \notag \\
  B_n(x) = \sum_{k=0}^n
    \binom{n}{k} B_k x^{n-k}, \notag \\
  B_n(0) = B_n(1) = B_n \quad (n &gt; 1).  \notag
\end{gather}

\begin{lemma}  \label{lem_berpoly}
Let $m,n \in \mathbb{N}$ and
\[ S_n(m) := \sum_{l=1}^m l^n. \]
Then we have
\[ S_n(m) = \frac{m^{n+1}}{n+1} + \frac{1}{2} m^n + \frac{1}{n+1}
           \sum_{k=1}^{n-1} \binom{n+1}{k+1} B_{k+1} m^{n-k}. \]
\end{lemma}

\begin{proof}
We have
\[ \begin{split}
  S_n(m) &amp;= \frac{1}{n+1} (B_{n+1}(m+1) - B_{n+1}(1))  \\
         &amp;= \frac{1}{n+1}(B_{n+1}(m) + (n+1)m^n - B_{n+1}) \\
         &amp;= \frac{1}{n+1} \biggl( \sum_{k=0}^{n+1}
           \binom{n+1}{k} B_k m^{n+1-k}
           + (n+1)m^n - B_{n+1} \biggr) \\
         &amp;= \frac{m^{n+1}}{n+1} + \frac{1}{2} m^n + \frac{1}{n+1}
           \sum_{k=1}^{n-1} \binom{n+1}{k+1} B_{k+1} m^{n-k}.
\end{split} \]
\end{proof}

We use the following lemmas in the proof of Theorem \ref{th_pos}.

\begin{lemma} \label{lem_phik}
Let $r, k \in \mathbb{N}$ with $r &gt; k$ and $x \ge 1$. We have
\[ \sum_{n \le x} n^r \varphi_k(n)  \le x^{r+1}. \]
\end{lemma}

\begin{proof}
We have
\begin{align*}
  \sum_{n \le x} n^r \varphi_k(n) &amp;= \sum_{n \le x} n^{r-k}
    \sum_{d \mid n} \mu(d) d^k \\
  &amp;= \sum_{d \le x} \mu(d) d^k \bigl(d^{r-k} + (2d)^{r-k} + \cdots +
    (d \lfloor x/d \rfloor)^{r-k} \bigr) \\
  &amp;= \sum_{d \le x} \mu(d) d^r \sum_{j \le x/d} j^{r-k} \\
  &amp;\le x^{r+1}.
\end{align*}
\end{proof}

\begin{lemma} \label{lem_nphin}
Let $r \in \mathbb{N}$ and $x &gt; e$. We have
\[ \sum_{n \le x} n^r \varphi(n) = \frac{x^{r+2}}{(r+2) \zeta(2)}
     + O(x^{r+1} (\log x)^{2/3} (\log \log x)^{4/3})
     \quad (\text{as $x \to \infty$}), \]
where the implied constant depends on $r$.
\end{lemma}

\begin{proof}
We can obtain the lemma by the estimate \cite{walfisz}
\[ \sum_{n \le x} \varphi(n) = \frac{x^2}{2 \zeta(2)} +
     O(x (\log x)^{2/3} (\log \log x)^{4/3}) \]
and the partial summation formula.
\end{proof}

\section{Proof of Theorem \ref{th_conv} and Theorem \ref{th_pos}}
\begin{proof}[Proof of Theorem \ref{th_conv}]
We have
\[ \begin{split}
   \sum_{j=1}^n \biggl( \frac{j}{\gcd(n,j)} \biggr)^a &amp;=
   \sum_{d \mid n} \frac{1}{d^a}
     \sum_{\substack{j=1 \\ \gcd(j,n)=d}}^n j^a \\
   &amp;= \sum_{d \mid n} \frac{1}{d^a}
     \sum_{\substack{k \le n/d \\ \gcd(k,n/d)=1}} (kd)^a =
   \sum_{d \mid n} \sum_{\substack{k \le n/d \\ \gcd(k,n/d)=1}} k^a.
   \end{split} \]
By Lemma \ref{lem_berpoly} we have
\begin{align*}
  \sum_{\substack{k \le N \\ \gcd(k,N)=1}} k^a &amp;=
    \sum_{k \le N} k^a \sum_{d \mid \gcd(k,N)} \mu(d) =
    \sum_{d \mid N} d^a \mu(d) \sum_{m \le N/d} m^a \\
  &amp;= \sum_{d \mid N} d^a \mu(d) \biggl( \frac{1}{a+1}
    \biggl( \frac{N}{d} \biggr)^{a+1} + \frac{1}{2} \biggl(
    \frac{N}{d} \biggr)^a + \\
  &amp;\qquad + \frac{1}{a+1} \sum_{k=1}^{a-1}
    \binom{a+1}{k+1} B_{k+1}
    \biggl( \frac{N}{d} \biggr)^{a-k} \biggr) \\
  &amp;= \frac{N^a}{a+1} \sum_{d \mid N} \mu(d) \frac{N}{d} +
    \frac{N^a}{2} \sum_{d \mid N} \mu(d) + \\
  &amp;\qquad + \frac{N^a}{a+1}
    \sum_{k=1}^{a-1} \binom{a+1}{k+1}
    B_{k+1} \sum_{d \mid N} \mu(d) \biggl(
    \frac{d}{N} \biggr)^k \\
  &amp;= N^a \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 +
    \frac{1}{a+1} \sum_{k=1}^{a-1}
    \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr)(N).
\end{align*}
Hence we obtain
\begin{align*}
  L_a(n) &amp;= n^a \sum_{j=1}^n \biggl( \frac{j}{\gcd(n,j)} \biggr)^a \\
  &amp;= \sum_{d \mid n} \biggl( \frac{n}{d} \biggr)^{2a} \cdot
    \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 +
    \frac{1}{a+1} \sum_{k=1}^{a-1}
    \binom{a+1}{k+1} B_{k+1} \varphi_k \biggr) (n/d) \cdot d^a \\
  &amp;= (M_a * \mathrm{Id}^a)(n).
\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{th_pos}]
By Lemma \ref{lem_phik}, Lemma \ref{lem_nphin} and
Theorem \ref{th_conv}, we have
\begin{align*}
  \sum_{n \le x} L_a(n) &amp;= \sum_{n \le x} (M_a \ast \mathrm{Id}^a)(n)
    = \sum_{d \le x} d^a \sum_{m \le x/d} M_a(m) \\
  &amp;= \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a}
    \biggl( \frac{1}{a+1} \varphi + \frac{1}{2} \tau_0 + \frac{1}{a+1}
    \sum_{k=1}^{a-1} \binom{a+1}{k+1}
    B_{k+1} \varphi_k \biggr) (m) \\
  &amp;= \frac{1}{a+1} \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a} \varphi(m)
    + O(x^{a+1}) + \\
  &amp;\qquad + \frac{1}{a+1} \sum_{k=1}^{a-1}
    \binom{a+1}{k+1}
    B_{k+1} \sum_{d \le x} d^a \sum_{m \le x/d} m^{2a} \varphi_k(m) \\
  &amp;= \frac{1}{a+1} \sum_{d \le x} d^a \biggl(
    \frac{1}{(2a+2) \zeta(2)} \biggl( \frac{x}{d} \biggr)^{2a+2} + \\
  &amp;\qquad + O \biggl( \biggl( \frac{x}{d} \biggr)^{2a+1} (\log (x/d))^{2/3}
    (\log \log (x/d))^{4/3} \biggr) \biggr) + \\
  &amp;\qquad + O(x^{a+1}) + O \biggl( \sum_{d \le x} d^a (x/d)^{2a+1} \biggr) \\
  &amp;= \frac{x^{2a+2}}{(a+1)(2a+2) \zeta(2)} \sum_{d \le x}
    \frac{1}{d^{a+2}} + O(x^{2a+1} (\log x)^{2/3} (\log \log x)^{4/3}) + \\
  &amp;\qquad + O(x^{2a+1}).
\end{align*}
This implies the theorem.
\end{proof}

\section{Proof of Theorem \ref{th_neg}}
\begin{proof}[Proof of Theorem \ref{th_neg}]
Since we have
\begin{align*}
  L_{-k}(n)&amp;=\sum_{j=1}^{n}\frac{1}{(\lcm(n,j))^{k}}=
    \frac{1}{n^{k}}\sum_{j=1}^{n}\frac{(\gcd(n,j))^{k}}{j^{k}}=
    \frac{1}{n^{k}}\sum_{d|n}d^{k}
    \sum_{\substack{ j=1\\ \gcd(j,n)=d}}^{n}
    \frac{1}{j^{k}}\\
  &amp;=\frac{1}{n^{k}}\sum_{d|n}d^{k}
    \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}}
    \frac{1}{i^{k}d^{k}}\\
  &amp;=\frac{1}{n^{k}}\sum_{d|n}
    \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}}\frac{1}{i^{k}},
\end{align*}
we obtain
\begin{align*}
  \sum_{n=1}^{\infty}L_{-k}(n)&amp;=
    \sum_{n=1}^{\infty}\frac{1}{n^{k}}\sum_{d|n}
    \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}}
    \frac{1}{i^{k}}=\sum_{d=1}^{\infty}\sum_{j=1}^{\infty}
    \frac{1}{j^{k}d^{k}}\sum_{\substack{i\leq j\\ \gcd(i,j)=1}}
    \frac{1}{i^{k}}\\
  &amp;=\zeta(k)\sum_{j=1}^{\infty} \frac{1}{j^{k}}
    \sum_{\substack{i\leq j\\ \gcd(i,j)=1}}\frac{1}{i^{k}}\\
  &amp;=\zeta(k)\sum_{n=1}^{\infty}\frac{1}{n^{k}}
    (\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1).
\end{align*}
Also we have
\[ \sum_{n=1}^{\infty}\frac{1}{n^{k}}
     \left(\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1 \right) =
     1+\frac{1}{2}\sum_{n=2}^{\infty}\frac{1}{n^{k}}
     \left(\sum_{\substack{ \gcd(i,j)=1\\ ij=n}}1 \right) \]
and the relation \cite[1.2.8]{titchmarsh}
\[ \sum_{n=1}^{\infty}\frac{1}{n^{k}}
     \left(\sum_{\substack{ \gcd(i,j)=1\\ ij=n}}1 \right)=
     \frac{\zeta(k)^{2}}{\zeta(2k)}. \]
Therefore we obtain
\[ \sum_{n=1}^{\infty} \frac{1}{n^{k}}
     \left(\sum_{\substack{i\leq j\\ \gcd(i,j)=1\\ ij=n}}1 \right)=
     1+\frac{1}{2} \biggl( \frac{\zeta(k)^{2}}{\zeta(2k)}-1 \biggr)=
     \frac{1}{2} \biggl(1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr). \]
This implies (\ref{series_lkn}).

By the relation
\[ \sum_{n \le x} L_{-k}(n) = \frac{\zeta(k)}{2} \biggl(
     1+\frac{\zeta(k)^{2}}{\zeta(2k)} \biggr) - \sum_{n &gt; x}
     L_{-k} (n), \]
the remaining task is to estimate the sum
$\sum_{n &gt; x} L_{-k} (n)$. We have
\begin{align*}
  \sum_{n&gt;x}L_{-k}(n)&amp;=\sum_{n&gt;x}\frac{1}{n^{k}}\sum_{d|n}
    \sum_{\substack{i\leq \frac{n}{d}\\ \gcd(i,\frac{n}{d})=1}}
    \frac{1}{i^{k}}=\sum_{d=1}^{\infty}
    \sum_{h&gt;\frac{x}{d}}\frac{1}{(hd)^{k}}
    \sum_{\substack{j\leq h\\ \gcd(j,h)=1}}\frac{1}{j^{k}}\\
  &amp;=\sum_{d=1}^{\infty}\sum_{h&gt;\frac{x}{d}}\frac{1}{(hd)^{k}}
    \sum_{j\leq h}\frac{1}{j^{k}}\sum_{\delta| \gcd(j,h)}\mu(\delta)\\
  &amp;=\sum_{d=1}^{\infty}\sum_{h&gt;\frac{x}{d}}\frac{1}{(hd)^{k}}
    \sum_{\delta|h}\sum_{m\leq\frac{h}{\delta}}
    \frac{\mu(\delta)}{m^{k}\delta^{k}}\\
  &amp;=\sum_{d=1}^{\infty}\frac{1}{d^{k}}\sum_{\delta=1}^{\infty}
    \sum_{l&gt;\frac{x}{d\delta}}\frac{\mu(\delta)}{l^{k}\delta^{2k}}
    \sum_{m\leq l}\frac{1}{m^{k}}\\
  &amp;=\sum_{q=1}^{\infty}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
    \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}\sum_{m\leq l}\frac{1}{m^{k}}\\
  &amp;=\sum_{q=1}^{\infty}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
    \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}
    \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)  \\
  &amp;= \sum_{q &lt; x} \frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
    \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}
    \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr) + \\
  &amp;\qquad + \sum_{q \ge x} \frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
    \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}
    \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)  \\
  &amp;=: S_1 + S_2,
\end{align*}
say. We have
\begin{align*}
S_1 &amp;= \sum_{q&lt;x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
  \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}
  \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\
&amp;=\sum_{q&lt;x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
  \biggl(\frac{\zeta(k)}{k-1}(x/q)^{-k+1}+O((x/q)^{-k})- \\
&amp;\qquad - \frac{(\frac{x}{q})^{-2k+2}}{(k-1)(2k-2)}+O((x/q)^{-2k+1})
  \biggr)\\
&amp;=\sum_{q&lt;x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
  \biggl(\frac{\zeta(k)}{k-1}(x/q)^{-k+1}+O((x/q)^{-k}) \biggr)\\
&amp;=\sum_{q&lt;x}\frac{1}{q^{k}}\sum_{d|q}\frac{\mu(d)}{d^{k}}
  \biggl(\frac{\zeta(k)}{k-1}(x/q)^{-k+1}+O((x/q)^{-k}) \biggr)\\
&amp;=\frac{\zeta(k)x^{-k+1}}{k-1}\sum_{q&lt;x}q^{-1}\sum_{d|q}
  \frac{\mu(d)}{d^{k}}+O \biggl(x^{-k}\sum_{q&lt;x}\sum_{d|q}
  \frac{|\mu(d)|}{d^{k}} \biggr)\\
&amp;=\frac{\zeta(k)x^{-k+1}}{k-1}\sum_{d&lt;x}\sum_{j&lt;\frac{x}{d}}j^{-1}
  \frac{\mu(d)}{d^{k+1}}+O(x^{-k+1})\\
&amp;=\frac{\zeta(k)x^{-k+1}}{k-1}\sum_{d&lt;x}\log \Bigl(\frac{x}{d} \Bigr)
  \frac{\mu(d)}{d^{k+1}}+O(x^{-k+1})\\
&amp;=\frac{\zeta(k)x^{-k+1}\log x}{(k-1)\zeta(k+1)}+O(x^{-k+1})
\end{align*}
and
\begin{align*}
&amp;\sum_{q\geq x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
  \sum_{l&gt;\frac{x}{q}}\frac{1}{l^{k}}
  \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\
&amp;\ll \sum_{q\geq x}\frac{1}{q^{2k}}\sum_{d\delta=q}d^{k}\mu(\delta)
  \sum_{l=1}^{\infty}\frac{1}{l^{k}}
  \biggl(\zeta(k)-\frac{l^{1-k}}{k-1}+O(l^{-k}) \biggr)\\
&amp;\ll\sum_{q\geq x}q^{-k}\sum_{d|q}\frac{|\mu(d)|}{d^{k}}\\
&amp;\ll x^{-k+1}.
\end{align*}
Therefore we obtain
\[ \sum_{n&gt;x}L_{-k}(n)=\frac{\zeta(k)x^{-k+1}\log x}{(k-1)\zeta(k+1)}
     +O(x^{-k+1}). \]
This completes the proof.
\end{proof}

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11N37.

\noindent \emph{Keywords: }
arithmetic function, lcm-sum function, least common multiple.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A018804} and
\seqnum{A051193}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 19 2013;
revised versions received November 14 2013; November 23 2013; December 19
2013.
Published in {\it Journal of Integer Sequences}, December 27 2013.

\bigskip
\hrule
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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\end{document}

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