# Find the limit $\lim\limits_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}$

This is a math competition problem for college students in Sichuan province, China. As the title, calculate the limit $$\lim_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}.$$ It is clear that the Dirichlet series $$\sum_{n=1}^\infty\frac{\sin n}{n^s}$$ is convergent for all complex number $$\Re s>0$$. Here we only consider the case of real numbers.

Let $$A(x)=\sum_{n\leq x}\sin n,$$ then we have that $$A(x)=\frac{\cos\frac{1}{2}-\cos([x]+\frac{1}{2})}{2\sin\frac{1}{2}},$$ here $$[x]$$ is the floor function. Obviously, $$A(x)$$ is bounded and $$|A(x)|\leq\frac{1}{\sin(1/2)}$$. Using Abel's summation formula, we have that $$\sum_{n=1}^\infty\frac{\sin n}{n^s}=s\int_1^\infty\frac{A(x)}{x^{s+1}}\,dx =\frac{\cos\frac{1}{2}}{2\sin\frac{1}{2}}-s\int_1^\infty\frac{\cos([x]+\frac{1}{2})}{x^{s+1}}\,dx.$$

The integral $$\int_1^\infty\frac{\cos([x]+\frac{1}{2})}{x^{s+1}}\,dx$$ or $$\int_1^\infty\frac{\cos([x])}{x^{s+1}}\,dx$$ is also convergent for $$s>-1$$ (am I right? use Dirichlet's test)

My question: Is there an easy way to prove $$\lim_{s\to0^+}\int_1^\infty\frac{\cos([x])}{x^{s+1}}\,dx= \int_1^\infty\lim_{s\to0^+}\frac{\cos([x])}{x^{s+1}}\,dx= \int_1^\infty\frac{\cos([x])}{x}\,dx.$$ If the above conclusion is correct, we have that $$\lim_{s\to0^+}\sum_{n=1}^\infty\frac{\sin n}{n^s}=\frac{\cos\frac{1}{2}}{2\sin\frac{1}{2}}.$$

More generally, consider the Mellin tranform $$g(s)=\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx$$, here $$f(x)$$ is continuous except integers and have left and right limit at integers.

If for $$s=0$$, the integral $$\int_1^\infty\frac{f(x)}{x}\,dx$$ is convergent, do we have that $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx\stackrel{?}=\int_1^\infty\frac{f(x)}{x}\,dx\,$$(In Jameson's book The prime number theorem, page 124, there is a Ingham-Newman Tauberian thereom, but the conditions of the theorem there are slightly different from here.)

If the condition is strengthened to for all $$s\geq-1/2$$, $$\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx$$ is convergent, is the following correct $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx\stackrel{?}=\int_1^\infty\frac{f(x)}{x}\,dx\,$$ If this is correct, is there a simple way to prove it?

(2022/3/24/21:53) If the Dirichlet integral $$\int_1^\infty\frac{f(x)}{x^s}dx$$ converges at $$s_0$$, then it converges uniformly in $$|\arg(s-s_0)|\leq\alpha<\frac{\pi}{2}$$ for any fixed $$0<\alpha<\frac{\pi}{2}$$, and thus $$\lim_{s\to s_0^+}\int_1^\infty\frac{f(x)}{x^s}dx=\int_1^\infty\frac{f(x)}{x^{s_0}}dx.$$

For the uniform convergence of Dirichlet integral, see Uniform convergence about Dirichlet integral $f(s):=\int_1^\infty\frac{a(x)}{x^s}\,dx =\lim\limits_{T\to\infty}\int_1^T\frac{a(x)}{x^s}\,dx$

• You just need that the integral of that cosine is convergent for all $s \geq 0$. In particular, it is bounded for all $s\geq 0$. So since it is multiplied by $s$, it will tend to $0$ as $s \to 0+$.
– Gary
Commented Mar 19, 2022 at 4:22
• Note that I said $s\geq 0$ and not $s>0$.
– Gary
Commented Mar 19, 2022 at 4:36
– Gary
Commented Mar 19, 2022 at 5:05
• I think the dominated convergence theorem does the trick right? Commented Mar 19, 2022 at 5:39
• @JackT: If $\int_1^\infty\left|\frac{\cos\left(\lfloor x\rfloor\right)}x\right|\,\mathrm{d}x$ were convergent, then we could apply Dominated Convergence. However, cancellation is important in the convergence of $\int_1^\infty\frac{\cos\left(\lfloor x\rfloor\right)}x\,\mathrm{d}x$, so I don't think we can apply Dominated Convergence.
– robjohn
Commented Mar 20, 2022 at 21:10

A Couple of Trigonometric Sums

First, we evaluate \newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}} \begin{align} S_n &=\sum_{k=1}^n\sin(k)\tag{1a}\\ &=\Im\left(\frac{e^{i(n+1)}-1}{e^i-1}\right)\tag{1b}\\ &=\Im\left(e^{in/2}\right)\frac{\sin\left(\frac{n+1}2\right)}{\sin\left(\frac12\right)}\tag{1c}\\ &=\sin\left(\frac n2\right)\frac{\sin\left(\frac{n+1}2\right)}{\sin\left(\frac12\right)}\tag{1d}\\ &=\frac{\cos\left(\frac12\right)-\cos\left(n+\frac12\right)}{2\sin\left(\frac12\right)}\tag{1e} \end{align} Explanation:
$$\text{(1a)}$$: definition
$$\text{(1b)}$$: apply Euler's Formula
$$\phantom{\text{(1b):}}$$ and the Formula for the Sum of a Geometric Series
$$\text{(1c)}$$: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\text{(1d)}$$: apply Euler's Formula
$$\text{(1e)}$$: $$\sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}2$$

Similarly, \begin{align} C_n &=\sum_{k=1}^n\cos\left(k+\frac12\right)\tag{2a}\\ &=\Re\left(\frac{e^{i(n+3/2)}-e^{i3/2}}{e^i-1}\right)\tag{2b}\\ &=\Re\left(e^{i(n+3)/2}\right)\frac{\sin\left(\frac{n}2\right)}{\sin\left(\frac12\right)}\tag{2c}\\ &=\cos\left(\frac{n+3}2\right)\frac{\sin\left(\frac{n}2\right)}{\sin\left(\frac12\right)}\tag{2d}\\ &=\frac{\sin\left(n+\frac32\right)-\sin\left(\frac32\right)}{2\sin\left(\frac12\right)}\tag{2e} \end{align} Explanation:
$$\text{(2a)}$$: definition
$$\text{(2b)}$$: apply Euler's Formula
$$\phantom{\text{(2b):}}$$ and the Formula for the Sum of a Geometric Series
$$\text{(2c)}$$: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\text{(2d)}$$: apply Euler's Formula
$$\text{(2e)}$$: $$\sin(x)\cos(y)=\frac{\sin(x+y)+\sin(x-y)}2$$

Estimating the Sum

Therefore, \begin{align} \sum_{k=1}^\infty\frac{\sin(k)}{k^s} &=\sum_{k=1}^\infty\frac{S_k-S_{k-1}}{k^s}\tag{3a}\\ &=\sum_{k=1}^\infty S_k\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag{3b}\\ &=\frac12\cot\left(\frac12\right)-\frac12\csc\left(\frac12\right)\sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag{3c}\\ &=\frac12\cot\left(\frac12\right)+O(s)\tag{3d} \end{align} Explanation:
$$\text{(3a)}$$: $$\sin(k)=S_k-S_{k-1}$$
$$\text{(3b)}$$: Summation by Parts
$$\text{(3c)}$$: apply $$(1)$$ and $$\sum\limits_{k=1}^\infty\left(\frac1{k^s}-\frac1{(k+1)^s}\right)=1$$
$$\text{(3d)}$$: $$\frac1{k^s}-\frac1{(k+1)^s}$$ is monotonic decreasing and $$1-\frac1{2^s}\le s$$
$$\phantom{\text{(3d):}}$$ $$(2)$$ says that $$\sup\limits_{n\ge0}\left|\sum\limits_{k=1}^n\cos\left(k+\frac12\right)\right|\le\csc\left(\frac12\right)$$
$$\phantom{\text{(3d):}}$$ Dirichlet says $$\left|\sum\limits_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\right|\le s\csc\left(\frac12\right)$$

The Requested Result

Estimate $$(3)$$ yields $$\bbox[5px,border:2px solid #C0A000]{\lim_{s\to0^+}\sum_{k=1}^\infty\frac{\sin(k)}{k^s}=\frac12\cot\left(\frac12\right)}\tag4$$

Clarification

More than one comment has shown that the bound on the sum $$\sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\tag5$$ given in $$\text{(3d)}$$ requires clarification.

Using the Generalized Dirichlet Convergence Test, as presented in this answer, we will set $$a_k=\cos\left(k+\frac12\right)\tag6$$ and $$b_k=\frac1{k^s}-\frac1{(k+1)^s}\tag7$$ In $$(2)$$, it is shown that $$\left|\,\sum_{k=1}^na_k\,\right|\le\csc\left(\frac12\right)\tag8$$ Since $$x^{-s}$$ is convex, \begin{align} &\overbrace{\left(\frac1{(k+1)^s}-\frac1{(k+2)^s}\right)}^{\large b_{k+1}}-\overbrace{\left(\frac1{k^s}-\frac1{(k+1)^s}\right)}^{\large b_k}\tag{9a}\\ &=2\left(\frac1{(k+1)^s}-\frac12\left(\frac1{k^s}+\frac1{(k+2)^s}\right)\right)\tag{9b}\\[4pt] &\le0\tag{9c} \end{align} Explanation:
$$\text{(9a)}$$: definition of $$b_k$$
$$\text{(9b)}$$: combine terms
$$\text{(9c)}$$: $$f\left(\frac{x+y}2\right)\le\frac{f(x)+f(y)}2$$ for convex $$f$$

Thus, $$b_k$$ decreases monotonically to $$0$$. Therefore, the total variation of $$b_k$$ is \begin{align} \sum_{k=1}^\infty|b_k-b_{k+1}| &=\sum_{k=1}^\infty(b_k-b_{k+1})\tag{10a}\\ &=b_1\tag{10b}\\[9pt] &=1-2^{-s}\tag{10c}\\[9pt] &=1-(1+1)^{-s}\tag{10d}\\[9pt] &\le1-(1-s)\tag{10e}\\[9pt] &=s\tag{10f} \end{align} Explanation:
$$\text{(10a)}$$: $$b_k$$ is monotonically decreasing
$$\text{(10b)}$$: telescoping sum
$$\text{(10c)}$$: evaluate $$b_1$$
$$\text{(10d)}$$: $$1+1=2$$
$$\text{(10e)}$$: Bernoulli's Inequality
$$\text{(10f)}$$: simplify

Applying the Generalized Dirichlet Convergence Test to $$(8)$$ and $$(10)$$, we get $$\left|\,\sum_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\,\right|\le s\csc\left(\frac12\right)\tag{11}$$

• Use Dirichlet's test: $|\sum_{k=1}^n\cos(k+\frac{1}{2})|\leq A$, $\sum_{k=1}^\infty|\frac{1}{k^s}-\frac{1}{k^{s+1}}|\leq 1$, then we can only get $\left|\sum\limits_{k=1}^\infty\cos\left(k+\frac12\right)\left(\frac1{k^s}-\frac1{(k+1)^s}\right)\right|\le A\cdot 1=A$, not $\frac{s}{2}\cdot\csc(1/2)$. Am I right? Maybe we can prove that $\sum_{k=1}^\infty\cos(k+\frac{1}{2})(\frac{1}{k^s}-\frac{1}{(k+1)^s})$ is uniform convergent on every closed interval $[0,M]$ and take limit.
– HGF
Commented Mar 21, 2022 at 9:04
• @HGF: $a_k=\cos\left(k+\frac12\right)$ and $b_k=\frac1{k^s}-\frac1{(k+1)^s}$. The partial sums of $a_k$ are uniformly bounded by $(2)$. $b_k$ is monotonically decreasing to $0$, therefore, the total variation of $b_k$ is $b_1\le s$. Remember, we are not summing against $\frac1{k^s}$ but $b_k=\frac1{k^s}-\frac1{(k+1)^s}$.
– robjohn
Commented Mar 21, 2022 at 9:33
• Ok, I misunderstood， thanks a lot.
– HGF
Commented Mar 21, 2022 at 10:49
• @robjohn Hi Rob. I hope that you are doing well. I have the same question as HGF had. Yes, $\frac1{k^\alpha}-\frac1{(k+1)^\alpha}<1-\frac1{2^s}<s$. But how can one assert that $\sum_{k\ge1}\left(\frac1{k^\alpha}-\frac1{(k+1)^\alpha} \right)=1<s$? That is obviously not true for $s<1$. Commented Apr 16, 2022 at 21:23
• @MarkViola: Please re-read my reply to HGF: $b_k=\frac1{k^s}-\frac1{(k+1)^s}$; we are not using $b_k=\frac1{k^s}$. Since $b_1=1-\frac1{2^s}\le s$, and $b_k$ is decreasing, the total variation of $b_k$ is at most $s$. The sum is less than the bound of the $a_k$'s times the variation of the $b_k$'s.
– robjohn
Commented Apr 16, 2022 at 23:27

Your claim is correct and the tool is integration by parts. The function $$f(x)=\cos(\lfloor x \rfloor + k)$$ admits a primitive $$F(x)$$ which is uniformly bounded, and after IP, the new integral involves integrating $$F(x)/(x^{s+2})$$. It behaves nicely and you may take limits.

IP also provides some natural conditions for your second question. For example, it suffices that $$f$$ has a primitive $$F$$ such that $$F(x)|< C(1+|x|^\alpha)$$ with $$\alpha<1$$.

Incidentally, doing repeated IP also implies that $$s\mapsto \sum_{n\geq 1} \frac{\sin(n)}{n^s}$$ extends to an entire function in the complex plane.

For you last question in the generality stated I don't know.

For $$\Re(s) >1$$ $$F(s)=\Gamma(s)\sum_{n\ge 1} \sin(n)n^{-s}=\sum_{n\ge 1}\int_0^\infty x^{s-1} \sin(n)e^{-nx}dx=\int_0^\infty x^{s-1} \Im(\frac{1}{e^{x-i}-1})dx$$ Next, note that $$F(s)- \Im(\frac{1}{e^{-i}-1})\Gamma(s)=\int_0^\infty x^{s-1} (\Im(\frac{1}{e^{x-i}-1})-\Im(\frac{1}{e^{-i}-1}) e^{-x})dx$$ converges and is continuous for $$\Re(s) >-1$$.

Whence $$\lim_{s\to 0}\sum_{n\ge 1} \sin(n)n^{-s}=\Im(\frac{1}{e^{-i}-1})=\frac{\cos(1/2)}{2\sin(1/2)}$$

• Has anyone historically studied the functional equation of the Dirichlet series $\sum_{n=1}^\infty\frac{\sin n}{n^s}$ or $\sum_{n=1}^\infty\frac{e^{in}}{n^s}$?
– HGF
Commented Mar 20, 2022 at 2:12
• Sure it is more or less the $1-s$ of the Hurwitz zeta. One way is to apply the residue theorem to $(e^{2i\pi }-1) \Gamma(s) \sum_{n\ge 1} \sin(n)n^{-s}=\int_C z^{s-1}\sum_{n\ge 1} \sin(n)e^{-nz}dz$ where $C$ is a Bromwich contour Commented Mar 20, 2022 at 2:13
• Did you mean to write $e^{x-i}$ instead of $e^{i+x}$? (line $2$) Commented Mar 20, 2022 at 18:43

I put Rugh's answer in detail here.

Let $$f$$ be an integrable function on $$[a,b]$$, $$g$$ a differentiable function on $$[a,b]$$ and let $$F(x)=\int_a^xf(t)\,dt,$$ then we have the integration by parts formula (see:Integration by Parts without Differentiation, Vicente Munoz,Mathematics Magazine , Vol. 85, No. 3 (June 2012), pp. 211-213): $$\int_a^bf(x)g(x)\,dx=g(x)\cdot F(x)\Big|_a^b-\int_a^bF(x)\,dg(x).$$ Here, all we need is that $$f$$ is integrable, not continuous.

Now, let $$F(x)=\int_1^x\cos([t]+a)\,dt,$$ Obviously, $$|F(x)|\leq C$$. or more generally, let $$F(x)=\int_1^x f(t)\,dt\quad\text{and assume that}\quad |F(x)|\leq C(1+|x|^\alpha),\quad \alpha<1.$$

Now $$\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx= \frac{F(x)}{x^{s+1}}\Big|_1^\infty+(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx=(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx.$$ It is clear that $$\int_1^\infty\left|\frac{F(x)}{x^{s+2}}\right|\,dx\leq\int_1^\infty\frac{C(1+|x|^\alpha)}{x^{2}}\,dx<+\infty.$$ Hence $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx=\lim_{s\to0^+}(s+1)\int_1^\infty\frac{F(x)}{x^{s+2}}\,dx=\int_1^\infty\frac{F(x)}{x^{2}}\,dx.$$ Since $$\int_1^\infty\frac{f(x)}{x}\,dx=\int_1^\infty\frac{dF}{x}=\int_1^\infty\frac{F(x)}{x^2}\,dx,$$ we have that $$\lim_{s\to0^+}\int_1^\infty\frac{f(x)}{x^{s+1}}\,dx=\int_1^\infty\frac{f(x)}{x}\,dx.$$

Let $$\operatorname{Li}_{s}(z)$$ be the polylogarithm function of order $$s$$.

For $$\Re(s) >0$$, we have $$\sum_{n=1}^{\infty} \frac{\sin(n)}{n^{s}} = \Im \operatorname{Li}_{s}(e^{i}).$$

And since $$\frac{\partial}{\partial z} \operatorname{Li}_{s+1}(z) = \frac{\operatorname{Li}_{s}(z)}{z}$$, we have$$\operatorname{Li}_{0}(z) = z\frac{\mathrm d }{\mathrm d z}\operatorname{Li}_{1}(z) = - z \frac{\mathrm d}{\mathrm d z} \ln(1-z) = \frac{z}{1-z} .$$

Therefore, $$\lim_{s \to 0^{+}}\sum_{n=1}^{\infty}\frac{\sin(n)}{n^{s}} = \Im \lim_{s \to 0^{+}} \operatorname{Li}_{s}(e^{i}) = \Im \, \frac{e^{i}}{1-e^{i}} = \Im \, \frac{i e^{i/2}}{2\sin \left(\frac{1}{2} \right)} = \frac{1}{2} \cot \left(\frac{1}{2} \right).$$

Similarly, we have$$\lim_{s \to 0^{+}}\sum_{n=1}^{\infty}\frac{\cos(n)}{n^{s}} = \Re \, \frac{e^{i}}{1-e^{i}} = - \frac{1}{2}.$$

UPDATE:

To prove that $$\lim_{s \to 0^{+}} \operatorname{Li}_{s}(e^{i}) = \operatorname{Li}_{0}(e^{i})$$, we need to show that $$\operatorname{Li}_{s}(e^{i})$$ is continuous at $$s=0$$.

For $$\Re(s) >0$$ and all $$z \in \mathbb{C}$$ except $$z$$ real and $$\ge 1$$, the polylogarithm has the Mellin transform representation $$\operatorname{Li}_{s}(z) = \frac{z}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{e^{x} -z} \, \mathrm dx.$$

If we integrate by parts, we get $$\operatorname{Li}_{s}(z) = \frac{z}{\Gamma(s+1)} \int_{0}^{\infty} \frac{x^{s}e^{x}}{\left(e^{x}-z\right)^{2}} \, \mathrm dx. \tag{1}$$

Since the Mellin transform defines an analytic function in the vertical strip where it converges converges absolutely (https://dlmf.nist.gov/1.14.iv), the above representation defines an analytic function in $$s$$, and thus a continuous function in $$s$$, in the half-plane $$\Re(s) >-1$$ for any fixed value of $$z$$ except $$z$$ real and $$\ge 1$$.

We can check to make sure that $$(1)$$ gives the correct expression for $$\operatorname{Li}_{0}(z)$$.

$$\operatorname{Li}_{0}(z) = \frac{z}{\Gamma(1)} \int_{0}^{\infty} \frac{e^{x}}{(e^{x}-z)^{2}} = - \frac{z}{e^{x}-z} \Bigg|^{\infty}_{0} =0 + \frac{z}{1-z} = \frac{z}{1-z}$$

• This is a wonderful answer, I have a question. For the fixed $z_0$, except for $z_0$ real and $z_0\geq1$, does the limit hold that $\lim\limits_{s \to s_0^{+}} \operatorname{Li}_{s}(z_0)=\operatorname{Li}_{s_0}(z_0)$? I saw the result in Wiki that $\operatorname{Li}_{s}(z)$ is analytic for all $z$ except $z$ real and $z\geq1$ and all complex number $s$ with $s\neq 1,2,3,\dots$. If this result holds, then there is nothing wrong with taking the limit, right?
– HGF
Commented Apr 1, 2022 at 2:41