Jan 1, 1980 - ... that Hall's result is best possible in the sense that the error estimate is given by $O(n^{-m-1}t^{\theta})$ ... $M(x)=\sum_{n\leq x}\mu(n)=O(x\delta_{A}(x))$ , ...... is as given in Lemma 8, $r>1$ and $\sigma(r)$ is the sum of all ...
J. Math. Soc.
Japan
Vol. 34, No. 1, 1982
Some sums involving Farey fractions I By Shigeru KANEMITSU1), R. SITA RAMA CHANDRA RAO and A. SIVA RAMA SARMA (Received Jan. 1, 1980) (Revised June 9, 1980)
1. Introduction. It is our aim in this paper to give some refinements of theorems proved by Hall [6] and the first author [9] on some sums involving Farey fractions. Let $F_{n}(n\in N)$ be the Farey series of order , that is, the set of all irreducible fractions between and arranged in the and 1 with denominators increasing order of magnitude: $F_{n}=\{h/k|0\leqq h\leqq k\leqq n, (h, k)=1\}$ ; for any term $h/k(0$
$t$
$u$
$\gamma$
$O(n^{-m-1}t^{\theta})$
$S_{n}(m)=2-\frac{m-1)}{m)n^{m}}\zeta(\zeta(+\frac{3\theta t}{\zeta(2)n^{4}}+o(\frac{1}{n^{m+1}})$
for
,
$m\geqq 3$
.
In section 3 we establish the following sharper results: , we have, as THEOREM 1. For each $\epsilon>0$
$ n\rightarrow\infty$
$S_{n}(2)=\frac{12}{\pi^{2}n^{2}}\{t+C+\frac{1}{2}-\frac{\zeta^{\prime}(2)}{\zeta(2)}\}+4U(n\underline{)t}_{-+O_{\epsilon}(\frac{tu^{1+\epsilon}}{n^{3}})}n^{3}$
where $t=t(n)=\log n,$ $ u=u(n)=\log$ log and $U(n)$ is given by (7). , we have, as THEOREM 2. For each $n$
$\epsilon>0$
$ n\rightarrow\infty$
$S_{n}(3)=\frac{2\zeta(2)\zeta^{-1}(3)}{n^{3}}+\frac{3\zeta^{-1}(2)}{n^{4}}\{t+C-\frac{1}{4}-\frac{\zeta^{\prime}(2)}{\zeta(2)}+2\zeta^{2}(2)c_{2}(n)\}$
$+\frac{12U}{n}(n)t5+O_{\epsilon}(\frac{tu^{1}}{n}5+\epsilon-)$
where for
$\lambda>1$
,
Sums involving Farey
(4)
fractions
127
$I$
,
$c_{\lambda}(n)=\sum_{m=1}^{\infty}\frac{\mu(m)}{m^{\lambda}}P_{1}(\frac{n}{m})$
have the same meanings as in Theorem 1. Now we fix some notation, recall some known results and establish necessary lemmas. Let $\mu(n)$ denote the M\"obius function, the Euler totient func$P_{n}(v)$ the periodic Bernoulli polynomial of degree , so that, tion and for $n\in N,$ $P_{1}(v)=\{v\}-1/2$ in particular, and $P_{2}(v)=\{v\}^{2}-\{v\}+1/6(\{v\}$ being the fractional part of ). Further, we write and
$t,$
$u$
$\phi(n)$
$n$
$v$
,
(5)
$E(x)=\sum_{n\leqq x}\phi(n)-\frac{x^{2}}{2\zeta(2)}$
(6)
$H(x)=\sum_{n\leqq x}\frac{\phi(n)}{n}-\frac{x}{\zeta(2)}$
(7)
$U(x)=\sum_{n\leqq x}\frac{\mu(n)}{n}P_{1}(\frac{x}{n})$
,
.
In order to make the error estimates as good as possible appearing in the lemmas below we shall use the following two best known estimates concerning $U(x)$ and the average of the M\"obius function due, respectively, to Saltykov [17] and Walfisz [20], although we do not need such deeper results for the proofs , as of our theorems: For each $\epsilon>0$
(8)
$ x\rightarrow\infty$
$U(x)=O^{\prime}\underline{\epsilon}\backslash \lambda_{\epsilon}(x))$
,
where (9)
$\lambda(x)=\lambda_{\epsilon}(x)\equiv t^{\gamma}u^{1+\text{\’{e}}}$
,
and (10)
as
$M(x)=\sum_{n\leq x}\mu(n)=O(x\delta_{A}(x))$
$ x\rightarrow\infty$
(11)
,
, where $\delta_{A}(x)=\exp(-At^{3/5}u^{-1/5})$
,
being a positive constant, not necessarily the same at each (Walfisz [20], p. 191 and also p. 181). $A$
occurrence
What we actually need is, as far as the order estimate is concerned, that for some $\alpha\in[0,1]$ , $\lambda(x)=O(x^{\alpha})$
and $\delta(x)=O(t^{-2}),$
$ x\delta(x)\uparrow\uparrow$
.
We make good use of the following LEMMA 1 (Euler-Maclaurin’s summation formula, cf. Rademacher [16], p. 14).
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S KANEMITSU, R. S. R. C. RAO and A. S. R. SARMA
Let
$a1$
$\sum_{n\leq x}\frac{\log n}{n^{s}}=-\zeta^{\prime}(s)-\frac{P_{1}(x)\log x}{x^{s}}-\frac{1}{(s-1)^{2}x^{S-1}}-\frac{t}{(s-1)x^{s- 1}}+o(\frac{1}{x^{s}})$
.
Now we have on the one hand, by (6) and partial summation $\sum_{n\leqq x}\frac{\phi(n)\log n}{n^{s+l}}=\frac{1}{(s-1)^{2}\zeta(2)}-\int_{1}^{\infty}\frac{H(t)}{t^{s+1}}dt+s\int_{1}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt$
(26)
$+\frac{H(x)\log x}{x^{s}}-\frac{1}{\zeta(2)(s-1)^{2}x^{s-1}}-\frac{\log x}{(s-1)\zeta(2)x^{s- 1}}$
$-s\int_{x}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt+o(\frac{\log x}{x^{s}})$
after simplification, where in the above the integrals converge in view of the trivial estimate $H(x)=O(\log x)$ . On the other hand, we have
Sums involving Farey
fractions
133
$I$
$\sum_{n\leqq x}\frac{\phi(n)\log n}{n^{s+l}}=\sum_{asx}\frac{\mu(d)\log d}{d^{s+l}}(\sum_{\delta\leqq x/d}\delta^{-s})$
$+\sum_{d\leqq x}\frac{\mu(d)}{d^{s+1}}$
(27)
$=S_{1}+S_{2}$
(
log ) $\delta$
$\sum_{\delta\leqq x/d}\delta^{-s}$
,
say. By using (24),
we find
$S_{1}=\sum_{d\leq x}\frac{\mu(d)\log d}{d^{s+l}}\{\zeta(s)-\frac{d^{s-1}}{(s-1)x^{s- 1}}-\frac{P_{1}\left(\begin{array}{l}x\\-d\end{array}\right)d^{s}}{x^{s}}+o((\frac{d}{x})^{s+1})\}$
(28)
$=\frac{\zeta(s)\zeta^{\prime}(s+1)}{\zeta^{2}(s+1)}-\frac{\zeta^{\prime}(2)}{(s-1)\zeta^{2}(s)x^{s- 1}}-\frac{1}{x^{s}}\sum_{d\leq x}\frac{\mu(d)\log d}{d}P_{1}(\frac{x}{d})$
$+O$ ( $x^{-s}$
log x) ,
where in the above we used $\sum_{a=1}d^{-s}\mu(d)\infty$
log
$d=\zeta^{\prime}(s)\zeta^{-2}(s)(s>1)$
$\sum_{d>x}d^{-S- 1}\mu(d)$
logd
$=O(x^{-s}\log x)$
and
that
.
Similarly, by using (25), we get (29)
$S_{2}=-\frac{\zeta^{\prime}(s)}{\zeta(s+1)}-\frac{\log x}{(s-1)\zeta(2)x^{s- 1}}+\frac{1}{(s-1)x^{s- 1}}(\frac{\zeta^{\prime}(2)}{\zeta^{2}(2)}-\frac{1}{(s-1)\zeta(2)})$
$-\frac{U(x)\log x}{x^{s}}+\frac{1}{x^{s}}\sum_{d\leqq x}\frac{\mu(d)\log d}{d}P_{1}(\frac{x}{d})+o(\frac{\log x}{x^{s}})$
where in the above we used obtain (30)
$\sum_{n=1}^{\infty}\mu(n)n^{-s}=1/\zeta(s)$
.
,
Thus by (27), (28) and (29), we
$\sum_{n\leqq x}\frac{\phi(n)\log n}{n^{s+l}}=\frac{\zeta(s)\zeta^{\prime}(s+1)}{\zeta^{2}(s+1)}-\frac{\zeta^{\prime}(s)}{\zeta(s+1)}-\frac{\log x}{(s-1)\zeta(2)x^{s- 1}}$
$-\frac{1}{(s-1)^{2}\zeta(2)x^{s-1}}-\frac{U(x)\log x}{x^{s}}+o(\frac{\log x}{x^{s}})$
Comparing the right sides of (26) and (30) and letting
$ x\rightarrow\infty$
,
.
we find, for
$s>1$
$\frac{1}{(s-1)^{2}\zeta(2)}-\int_{1}^{\infty}\frac{H(t)}{t^{s+1}}dt+s\int_{1}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt$
(31)
$=\frac{\zeta(s)\zeta^{\prime}(s+1)}{\zeta^{2}(s+1)}-\frac{\zeta^{\prime}(s)}{\zeta(s+1)}$
.
Again comparing the right sides of (26) and (30) and on using (31) together
134
S. KANEMITSU, R. S. R. C. RAO and A. S. R. SARMA
with (18), we deduce (32)
.
$\int_{x}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt=o(\frac{\log x}{x^{s}})$
According to (31) and (32), to prove the lemma it is sufficient to show that
for
$s>1$
(33)
$\int_{1}^{\infty}\frac{H(t)}{t^{s+1}}dt=_{\overline{s\zeta}(s1)}^{\zeta(}\frac{s)}{+}-\frac{1}{(s-1)\zeta(2)}$
.
In fact, on using (6) and partial summation, we get $\sum_{n\leqq x}\frac{\phi(n)}{n^{s+1}}=\frac{s}{(s-1)\zeta(2)}-\frac{1}{(s-1)\zeta(2)x^{s-1}}+s\int_{1}^{x}\frac{H(t)}{t^{s+1}}dt+o(\frac{\log x}{x^{s}})$
in the above, we get (33) in view of This completes the proof of Lemma 6.
By letting $(s>1)$
.
$ x\rightarrow\infty$
REMARK 2. Since
the integrals
$H(t)=O(\log t)$ ,
.
$\sum_{n=1}^{\infty}\phi(n)/n^{s+1}=\zeta(s)/\zeta(s+1)$
and
$\int_{1}^{\infty}\frac{H(t)}{t^{s+1}}dt$
$\int_{1}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt$
both converge absolutely and uniformly on every compact subset of the halfplane $\{s\in C|{\rm Re} s>0\}$ and hence define analytic functions. Thus by Lemma 6, (33) and analytic continuation we obtain: For any $s\in C$ with , ${\rm Re} s>0$
$\int_{1}^{\infty}\frac{H(t)\log t}{t^{s+l}}dt=\frac{\zeta(s)\zeta^{\prime}(s+1)}{s\zeta^{2}(s+1)}-\frac{\zeta^{\prime}(s)}{s\zeta(s+1)}+\frac{\zeta(s)}{s^{2}\zeta(s+1)}-\frac{1}{(s-1)^{2}\zeta(2)}$
$\int_{1}^{\infty}\frac{H(t)}{t^{s+1}}dt=\frac{\zeta(s)}{s\zeta(s+1)}-\frac{1}{(s-1)\zeta(2)}$
so
that, in particular,
,
$\int_{1}^{\infty}H(t)t^{-2}dt=\zeta^{-1}(2)(C-1-\zeta^{-1}(2)\zeta^{\prime}(2))$
LEMMA 7. There exist positive constants and any $r\in N$ $c$
log
$c$
and
$r\leqq\sum_{p\leqq r}\frac{\log p}{p}(\log\frac{r}{p})^{\alpha}\leqq d$
where the sum ranges over all primes PROOF. It is well-known that
$\leqq r$
.
such that
$d$
log
$r$
log
$P$
log
$\frac{r}{p}=\int_{0}^{r}\theta(t)t^{-1}dt=O(r)$
which in turn yields by partial summation $\sum_{p\leqq r}$
log
$\alpha\in[0,1]$
.
Hence $\sum_{p\leqq r}$
for any
,
$\theta(x)\equiv\sum_{pp^{p\leqq x}rime}\log P=O(x)$
[8], Theorem 414).
,
$p(\log\frac{r}{p})\frac{1}{p}=O(\log r)$
.
,
(Hardy and Wright
Sums involving Farey fractions
This together with the known
Theorem
425)
shows that for
135
$I$
$\sum_{p\xi r}(\log p)/p=\log r+O(1)$
(Hardy and Wright [8],
$\alpha\in[0,1]$
$\sum_{p\leqq r}\frac{\log p}{p}(\log\frac{r}{p})^{\alpha}\leqq\sum_{p\leqq r/e}\frac{\log p}{p}(\log\frac{r}{p})+\sum_{r/e0$
.
$T(x)=\sum_{k\leq x}t(k)$
.
Then
$ x\rightarrow\infty$
$T(x)=\frac{x}{\zeta(x)}\{\log x-1-\frac{\zeta^{\prime}(2)}{\zeta(2)}\}+H(x)$
log
$x+O_{\epsilon}(tu^{1+\epsilon})$
.
PROOF. Firstly, we note that (34)
$\alpha(k)\equiv-\frac{k}{\phi(k)}\sum_{d|k}\frac{\mu(d)\log d}{d}=\sum_{p|k}\frac{\log p}{p-1}$
,
the extreme sum on the right extending over all the distinct prime factors of . We remark parenthetically that Davenport [3] was the first to discuss systematically the function . (34) follows on noting that the functions involved on both sides are additive and coincide with each other at arbitrary prime powers. Secondly, we note that by (17), (18), Lemma 4 and the fact that $k$
$\alpha(k)$
$\delta_{A}(t)$
$=O((\log t)^{-2})$
as
$ t\rightarrow\infty$
$\int_{1}^{x}\frac{H(t)}{t}dt=O(1)+\int_{1}^{x}\frac{E(t)}{t^{2}}dt+o(\int_{3}^{x}\frac{\delta_{A}(t)}{t}dt)$
(35)
$=O(1)$
.
Now by (34) we obtain (36)
$T(x)=\sum_{k\leqq x}t(k)=\sum_{k\leqq x}\frac{\phi(k)\log k}{k}+\sum_{k\leqq x}\frac{\phi(k)\alpha(k)}{k}=S_{3}+S_{4}$
say. By the theorem of partial summation and (35), we have (37)
$S_{3}=\frac{x\log x}{\zeta(2)}-\frac{x}{\zeta(2)}+H(x)$
log $x+O(1)$
Further by (34) $S_{4}=\sum_{p\delta\leqq x}\frac{\phi(p\delta)\log p}{p\delta p-1}$
$=_{p}F_{\leq x}\frac{\phi(\delta)\log p}{\delta p}+\sum_{p\delta\leqq x,p^{1}\delta}\frac{\phi(\delta)}{\delta}\frac{\log p}{p(p-}1\overline{)}t$
.
,
S. KANEMITSU, R. S. R. C. RAO and A. S. R. SARMA
136
for each Now using successively (6), Lemma 5, the fact that (which is an easy consequence of (8) and (18)) and Lemma 7, we find $H(x)=O_{\epsilon}(\lambda_{\epsilon}(x))$
$\epsilon>0$
$S_{4}=\frac{x}{\zeta(2)}(\sum_{p}\frac{\log p}{p^{2}}+o(\frac{\log x}{x}))+O_{\epsilon}(\sum_{p\leqq x}\frac{\log p}{p}(\log\frac{x}{p})^{\gamma}(\log\log\frac{x}{p})^{1+\epsilon})$
$+\frac{x}{\zeta(2)}(\sum_{p}\frac{\log p}{p^{2}(p^{2}-1)}+o(\frac{\log x}{x^{3}}))+O_{\epsilon}(\lambda_{\epsilon}(x))$
\langle 38)
$=-\frac{\zeta^{\prime}(2)}{\zeta^{2}(2)}x+O_{\epsilon}(tu^{1+\epsilon})$
.
Collecting (36), (37) and (38), we conclude the assertion. , write and LEMMA 9. For $x\geqq 3$
$1\leqq Q\leqq x$
\langle 39)
$U(x, Q)=\sum_{n\leqq Q}\frac{\mu(n)}{n}P_{1}(\frac{x}{n})$
Then
for
each
.
$\epsilon>0$
(40)
$U(x, Q)=O_{\epsilon}(\lambda_{\epsilon}(x))$
is given by (9). and , where uniformly in Before establishing Lemma 9 we shall explain the meaning of . Consider the polynomial $f(y)=\alpha_{1}y+\cdots+\alpha_{n+1}y^{n+1}$ part of whose coefficients are rational, $(\nu=s+2, \cdots , 3s;1\leqq s\leqq(n+1)/3)$ and by say we mean the determinant $x$
$Q$
$\lambda_{\epsilon}(x)$
$\gamma$
$\Delta_{s}$
$\alpha_{\nu}=a_{\nu}/q$
det
$(\left(\begin{array}{l}s+i+j\\i\end{array}\right)a_{s+i+j})_{1\leqq i,j\leqq s}$
of order
$s$
.
Assume that if
is a fixed number in
$\delta$
the interval $0\gamma_{1}$
and the constants
$C_{1}$
and
$(\Delta_{s}, q)=1$
, then
,
$C_{2}$
depend possibly on , . It is known due to , thus yielding the $\delta$
we dePne With these Korobov [10] that (41) is valid for the choice and $\gamma=2/3$ . best known value PROOF OF LEMMA 9. First let be an integer. If $Q\leqq\exp(B\lambda(x)),$ being a constant depending on and , then (40) follows trivially. If exp $
