# Does $\sum_{n=1}^\infty \frac{\cos{(\sqrt{n})}}{n}$ converge?

The series is: $$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$

Considering it isn't always positive, I replace $$\frac{\cos{\sqrt{n}}}{n}$$ with its absolute value and I find that: $$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac{\cos{2\sqrt{n}}}{2n}$$ if $$\sum_{n=1}^\infty \vert\frac{\cos{\sqrt{n}}}{n}\vert$$ converges, then $$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$$ converges.Using Comparison test,we can draw the conclusion that $$\sum_{n=1}^\infty\frac{1}{2n}$$ converges , which is impossible. So I get that $$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$$ absolutely diverges. But I can't figure out whether $$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$$ converges or not.

I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=\sum_{k=1}^n \cos{\sqrt{k}}$$ is bounded.

(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)

• If you approximate the sum by an integral, then you have $$\sum_{n=1}^m\,\frac{\cos(\sqrt{n})}{n}\approx \int_1^m\,\frac{\cos(\sqrt{x})}{x}\,\text{d}x=2\,\text{Ci}(\sqrt{m})-2\,\text{Ci}(1)\,,$$ where $\text{Ci}$ is the cosine integral $$\text{Ci}(t)=-\int_t^\infty\,\frac{\cos(s)}{s}\,\text{d}s\,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$\lim_{m\to\infty}\,\text{Ci}(\sqrt{m})=0\,.$$ Oct 9, 2018 at 11:27
• Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$. Oct 9, 2018 at 11:32
• Nov 9, 2020 at 19:44

We show that $$N\to\sum_{n = 1}^N {\frac{\cos(\sqrt{n})}{n}}$$ is a Cauchy sequence and therefore the given series is convergent.

Hint. By using the MVT prove that for $$n\geq 1$$ and for all $$x\in [n,n+1)$$ $$\left|\frac{\cos(\sqrt{n})}{n}-\frac{\cos(\sqrt{x})}{x}\right|\leq \frac{1}{n^{3/2}}.$$ Then for $$M>N\geq 1$$, \begin{align} &\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}} - \int_N^{M + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} \right|\\ &=\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}} - \sum_{n = N}^M {\int_n^{n + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} } \right|\\ &\leq\sum_{n = N}^M \int_n^{n + 1}\left|{\frac{\cos(\sqrt{n})}{n}} - { {\frac{{\cos (\sqrt{x} )}}{x}} } \right|dx \le \sum_{n = N}^M \frac{1}{n^{3/2}}.\end{align} Note that since $$\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$$ is convergent then $$\lim_{M,N\to +\infty} \sum_{n = N}^M \frac{1}{n^{3/2}}=0.$$ Moreover $$\int_1^{+\infty}\frac{{\cos (\sqrt{x} )}}{x}dx=2\int_1^{+\infty}\frac{{\cos (u )}}{u}\, du$$ where the last integral is convergent, and we have that $$\lim_{M,N\to +\infty}\int_N^{M+1}\frac{{\cos (\sqrt{x} )}}{x}dx=0.$$

• So ultimately the absolute-term series is less than $\zeta(3/2)$ and thus the original series converges? Oct 9, 2018 at 11:49
• Not sure how the first inequality can lead to the second one. Oct 9, 2018 at 11:50
• @Szeto Is it clear now? Oct 9, 2018 at 11:55
• Thanks for your excellent proof! Oct 9, 2018 at 12:55

Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $$1/3$$.

n = 1
s = 0
s_min = s_max = -0.33

while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1


Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.

     n             s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)

• A quick estimate suggest you roughly need $\sim 10^{2k}$ terms to get convergence to $k$ digits. Oct 9, 2018 at 12:01
• Letting $a(k) = \frac{\cos \left(\sqrt{k}\right)}{k}$ Mathematica gives $s=\text{NSum[a[k], {k, 1, \infty}]} = -0.330688$ Oct 9, 2018 at 14:10
• Numeric inverse Laplace: $\gamma +\mathcal{L}_s^{-1}\left[\frac{\psi \left(1+s^2\right)}{s}\right](1)\approx -0.33068768045214513116891754615748$ where $\psi$ is PolyGamma and $\gamma$ is EulerGamma. Oct 24, 2018 at 17:42

Notations: $$\lfloor x \rfloor$$ is the floor function, $$\{x\}$$ is the fractional part of $$x$$ so that $$x=\lfloor x\rfloor + \{x\}$$.

Applying partial summation with $$f(x)=\frac{\cos(\sqrt x)}x$$, \begin{align} \sum_{n=1}^N \frac{\cos(\sqrt n)}n&=\int_{1-}^N f(x)d\lfloor x \rfloor \\ &=f(x)\lfloor x\rfloor \Big\vert_{1-}^N-\int_{1-}^N f'(x)\lfloor x\rfloor dx\\ &=f(N)(N-\{N\})-\int_1^N xf'(x)dx+\int_1^N\{x\}f'(x)dx\\ &=Nf(N)-f(1)-\int_1^Nxf'(x)dx+\int_1^N\{x\}f'(x)dx+f(1)-\{N\}f(N). \end{align} From integration by parts, the sum of first three terms is $$\int_1^{N}f(x)dx$$ Thus, we have the following as $$N\rightarrow\infty$$, $$\sum_{n=1}^{\infty}\frac{\cos(\sqrt n)}n=\int_1^{\infty}f(x)dx+\int_1^{\infty}\{x\}f'(x)dx+\cos 1.$$ It is easy to see that the integrals converge.

• @ i707107 I have derived the same final formula, but without using the differential $d\lfloor x\rfloor$ which looks strange to me. Could you please explain its meaning. Oct 10, 2018 at 13:06
• $d\lfloor x\rfloor$ means that you take the jump amount whenever there is a jump. The jumps of $\lfloor x \rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia. Oct 11, 2018 at 2:13
• Also, you can refer to Riemann-Stieltjes integral. Oct 11, 2018 at 2:14

This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $$d \lfloor x \rfloor$$.

Letting

$$c_{k} = c(k) = \frac{\cos(\sqrt{k})}{k}$$

we attempt to find an integral representation for the partial sum

$$s_n = \sum_{k=1}^n c_{k}$$

The formulas for partial summation are

$$\sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + \sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$

$$A_{k} = \sum_{i=1}^k a_{i}$$

Letting $$a_{k}=1, b_{k} = c_{k}$$ we have $$A_{k} = k$$.

Now comes the trick which introduces an integral: we have

$$(b_{k}-b_{k+1}) = - \int_{k}^{k+1} c'(x)\;dx$$

and, what's more, the factor $$k$$ can be incorporated in the integral:

$$k (b_{k}-b_{k+1}) = - k \int_{k}^{k+1} c'(x)\;dx = - \int_{k}^{k+1} \lfloor x\rfloor c'(x)\;dx$$

Where $$\lfloor x\rfloor$$ is the floor function.

Hence, using

$$\lfloor x\rfloor = x -\{x\}$$

where $$\{x\}$$ is the fractional part of $$x$$, the partial sum becomes

$$s_{n} = n c_n -\sum_{k=1}^{n-1} \int_{k}^{k+1} \lfloor x\rfloor c'(x)\;dx \\ = n c_n -\int_{1}^{n} x c'(x)\;dx + \int_{1}^{n} \{x\} c'(x)\;dx$$

By partial integration of the first integral the term $$n c_{n}$$ drops out and we get

$$s_{n} = \cos(1) +\int_{1}^{n} c(x)\;dx + \int_{1}^{n} \{x\} c'(x)\;dx$$

The first integral can be solved explicitly

$$\int_{1}^{n} c(x)\;dx = 2 \text{Ci}\left(\sqrt{n}\right)-2 \text{Ci}(1)$$

Where $$\text{Ci}$$ is the integral cosine.

The second integral is absolutely convergent and can be very roughly estimated thus (notice that $$0\le \{x\} \lt 1$$)

$$|i_2| = |\int_{1}^{n} \{x\} c'(x)\,dx| <= \int_{1}^{n}| \{x\} c'(x)|\,dx \\ <= \int_{1}^{n} |\{x\}| |c'(x)|\,dx <=\int_{1}^{n} | c'(x) | \,dx$$

Now

$$| c'(x) | = |\frac{\cos \left(\sqrt{x}\right)}{x^2}+\frac{\sin \left(\sqrt{x}\right)}{2 x^{3/2}}| \\ \leq | \frac{\sin \left(\sqrt{x}\right)}{2 x^{3/2}}| +|\frac{\cos \left(\sqrt{x}\right)}{x^2}| \leq \frac{1}{2 x^{3/2}}+\frac{1}{x^2}$$

Hence

$$| i_2| <=\int_1^n \left(\frac{1}{2 x^{3/2}}+\frac{1}{x^2}\right) \, dx = 2-\frac{\sqrt{n}+1}{n}\tag{*}$$

Since $$\lim_{n\to \infty } \, \text{Ci}\left(\sqrt{n}\right)= 0$$ the limit of the partial sum is given by

$$s = \cos(1) - 2 \text{Ci}(1) + \lim_{n\to \infty } \,i_2$$

Observing (*) $$s$$ remains finite as $$n\to\infty$$ hence the original sum is convergent. QED.

Remark

A more sophisticated study if the integral $$i_2$$ might provide better numerical bounds, and even a closed form.

This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.

Step 1: Transformation to an alternating sum

The partial sum in question is

$$s_n = \sum_{k=1}^n c_k\tag{1}$$

where

$$c_k = \frac{\cos(\sqrt{k})}{k}\tag{2}$$

Now collecting all subsequent summands with the same sign we can write

$$s_n = \sum_{m=0}^M (-1)^m f_m\tag{3}$$

where

$$f_0 = \sum_{k=1}^{\lfloor z_{1}\rfloor} c_{k}\tag{4a}$$

$$f_{m\ge 1} = (-1)^m \sum_{k={\lceil z_{m}\rceil}}^{\lfloor z_{m+1} \rfloor} c_{k}\tag{4b}$$

is the sum of terms between two adjacent roots $$k=z_m$$ and $$k=z_{m+1}$$ of $$c(k) = 0$$, and $$M$$ is some number depending on $$n$$ which can in principle be specified but it is normally not necessary as it goes to $$\infty$$ together with $$n$$

Then, by an appropirate choice of the factor $$(-1)^m$$, all $$f_m$$ can be chosen to be positive quantities.

Step 2: Application of Dirichlet's criterion

Now we can apply the Dirichlet criterion to (3). This reads now as follows: $$s_n$$ is convergent if

(1) $$f_m \to 0$$ for $$m\to\infty$$
(2) $$f_m$$ is monotonous

We have for $$m\ge1$$

$$|f_{m}| \le g(m)$$

where we have dropped the $$\cos$$ and have defined

$$g(m) = \sum_{k={\lceil z_{m}\rceil}}^{\lfloor z_{m+1} \rfloor} \frac{1}{k} =H_{\lfloor z_{m+1} \rfloor}-H_{\lceil z_{m}\rceil-1}\tag{5}$$

Here $$H_n$$ ist the harmonic number.

Now the zeroes of $$c_{k}$$ are given by

$$z_m = \left(\pi(m-\frac{1}{2})\right)^2\tag{6}$$

Dropping Floor and Ceiling in $$g(m)$$ defines

$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$

Letting $${\lfloor x \rfloor} \to x-1,{\lceil x \rceil} \to x+1$$ defines another function

$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$

and we have the inequality

$$g_1(m) \lt g(m) \lt g_0(m)$$

Asymptotically this leads, up to $$O(\frac{1}{m^2})$$, to the inequality

$$\frac{2}{m}- \frac{1}{\pi^2 m^2}< g(m) <\frac{2}{m}+\frac{1}{\pi^2 m^2}$$

This shows that $$f_m$$ goes to zero.

(to be continued)

## Outline

I haven't fully thought out this answer but leaving it here as a response to a comment on a duplicate question.

At the cost of introducing a constant $$C_N$$ we can start taking the sum at a large index $$N$$. $$\sum_{n=1} \cos (\sqrt n) = C_N +\sum_{n=N} \cos (\sqrt n)$$

Define $$x_n = \sqrt n$$ so have $$\sum_{n=N} \cos (\sqrt n) / n = \sum_{n=N} \cos (x_n) / x_n^2$$.

Defining $$\Delta x_n = x_{n+1} - x_{n}$$ we have $$\Delta x_n \approx \frac 1 {2 \sqrt n} = \frac 1 {2 x_n}$$ via first order taylor expansion of $$x(n+1) \approx x(n) + x'(n)\cdot 1$$

Substituting into the sum, we have $$\sum_{n=N} \cos (\sqrt n) / n \approx 2 \sum_{n=N}\cos (x_n) / x_n \Delta x_n$$. The right hand side is a Riemann / Darboux integral for $$2 \int_{\sqrt N} \frac {\cos x} x dx$$ and the integral is finite.

• That does make sense now Feb 24, 2021 at 13:33