1. ## Infinite series (4)

Prove that $\displaystyle \sum_{n=1}^{\infty} \left(1 + \frac{1}{2} + \cdots + \frac{1}{n} \right) \frac{\sin n}{n}$ is convergent.

2. can we use the inequality $\displaystyle \sum_{k=1}^{n} \frac{1}{k} < n$ ( the equality holds only when n = 1 ) ?

the next step we have to go is just considering whether $\displaystyle \sum_{n=1}^{\infty} {\sin n}$ converges

3. Originally Posted by simplependulum
can we use the inequality $\displaystyle \sum_{k=1}^{n} \frac{1}{k} < n$ ( the equality holds only when n = 1 ) ?

the next step we have to go is just considering whether $\displaystyle \sum_{n=1}^{\infty} {\sin n}$ converges
the comparison test is valid for series with positive terms only. so it doesn't work here. also, even if we could use the test, it still wouldn't give us anything because $\displaystyle \sum \sin n$ is divergent.

4. Maybe use the integral test and we know that $\displaystyle \text{Si}(\infty) = \pi/2$.

5. If we indicate with...

$\displaystyle H_{n} = \sum_{k=1}^{n} \frac{1}{k}$ (1)

... the 'armonic number' of index n, the 'infinite series' is...

$\displaystyle S= \sum_{n=1}^{\infty} H_{n} \frac{\sin n}{n}$ (2)

Remembering that the generating function of armonic numbers is...

$\displaystyle \sum_{n=1}^{\infty} H_{n}\cdot x^{n}= - \frac {\ln (1-x)}{1-x}$ (3)

... we derive in two steps...

$\displaystyle \sum_{n=1}^{\infty} H_{n}\cdot x^{n-1}= - \frac{\ln (1-x)}{x\cdot (1-x)}$ (5)

$\displaystyle \sum_{n=1}^{\infty} H_{n}\cdot \frac{x^{n}}{n} = -\int_{0}^{x} \frac{\ln (1-\xi)}{\xi\cdot (1-\xi)}\cdot d\xi = \varphi(x)$ (6)

Now we use the identity...

$\displaystyle \sin n = \frac{e^{in}-e^{-in}}{2i}$ (7)

... and (6) to arrive at the result...

$\displaystyle \sum_{n=1}^{\infty} H_{n}\cdot \frac{\sin n}{n} = \frac{1}{2i} \cdot \sum_{n=1}^{\infty} \frac{H_{n}}{n}\cdot (e^{in}-e^{-in}) = \frac{1}{2i}\cdot \{\varphi(e^{i}) - \varphi(e^{-i})\}$ (8)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Investigate the convergence of series $\displaystyle b_n=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n)\frac{\sin n}{n}$.

It is well known that $\displaystyle (1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n)=\gamma+O(\frac{1}{n})$, where $\displaystyle \gamma$ is the Euler-Mascheroni constant.

Obviously series $\displaystyle {b_n}$ converges since $\displaystyle b_n=\gamma\frac{\sin n}{n}+O(\frac{1}{n})\frac{\sin n}{n}$, knowing $\displaystyle \int^{\infty}_{1} \frac{\sin x}{x}dx$ exists.

Now investigate the convergence of series $\displaystyle c_n=\ln n \frac{\sin n}{n}$, leaving to check if the following integral exists
$\displaystyle \int^{\infty}_{1} \frac{\ln x}{x}\sin x dx$. Integral by part, it is easy to know it exists.

Therefore, $\displaystyle a_n=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})\fr ac{\sin n}{n}=b_n+c_n$ converges.

7. Luobo,

What you did is correct.
But it doesn't help for the present problem, since we want to prove if the series $\displaystyle \sum a_n$ is convergent, not the sequence $\displaystyle a_n$.

chisigma, how do you prove that $\displaystyle \varphi(e^i)-\varphi(e^{-i})$ converges ?

8. Moo,

I indeed is talking about series. For clarity, it is rearranged below:

(1) Series $\displaystyle \{b_n\}$ is convergent.
$\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n$ { Note: $\displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right)$ }
The first term converges knowing $\displaystyle \int^{\infty}_{1}\frac{\sin x}{x}\,dx$ is bounded, the second term converges even more obviously.

(2) Series $\displaystyle \{c_n\}$ is also convergent.
$\displaystyle \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n$
It converges since $\displaystyle \int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx$ is bounded.
$\displaystyle \left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|$$\displaystyle \leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty (3) Series \displaystyle \{a_n\} is convergent \displaystyle \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n 9. Originally Posted by Moo ... chisigma, how do you prove that \displaystyle \varphi(e^i)-\varphi(e^{-i}) converges?... ... all right!... with little patience [or even using Wolfram Integrator ...] you find that... \displaystyle \varphi (x)= - \int_{0}^{x} \frac{\ln(1-\xi)}{\xi(1-\xi)} d\xi= \frac{1}{2} \ln^2 (1-x) + Li_{2} (x) (1) ... where \displaystyle Li_{2} (*) is the Polylogarithm function of order two, defined in the complex plane as... \displaystyle Li_{2} (z)= \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} (2) Now the series (2) converges for \displaystyle |z|=1 so that \displaystyle \varphi (e^{i}) and \displaystyle \varphi (e^{-i}) both exist... Kind regards \displaystyle \chi \displaystyle \sigma 10. Following the same procedure we arrive to demonstrate that is... \displaystyle \sum_{n=1}^{\infty} H_{n} \frac{\cos n}{n} = \frac{1}{2} \{\varphi (e^{i}) + \varphi(e^{-i})\} (1) These are interesting results that are not isolated, because the same procedure can be used to arrive to even better results ... Kind regards \displaystyle \chi \displaystyle \sigma 11. An example... from the well known expression... \displaystyle \sum_{n=1}^{\infty} x^{n-1}= \frac{1}{1-x} (1) ... we derive... \displaystyle \sum_{n=1}^{\infty} \frac{x^{n}}{n} = \int_{0}^{x} \frac{d\xi}{1-\xi} = -\ln (1-x) (2) Now from (2) and the identity... \displaystyle \sin n= \frac{e^{in}-e^{-in}}{2i} (3) ... we derive... \displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n} = \frac{\ln (1-e^{-i}) - \ln (1-e^{i})}{2i}= \tan^{-1} \frac {\sin 1}{1-\cos 1} = 1.07079632679\dots (4) ... and from (2) and the identity... \displaystyle \cos n= \frac{e^{in}+e^{-in}}{2} (5) ... we derive... \displaystyle \sum_{n=1}^{\infty} \frac{\cos n}{n} = \frac{\ln (1-e^{-i}) + \ln (1-e^{i})}{2}= \ln \sqrt {2 (1-\cos 1)} = -.042019505825\dots (6) Kind regards \displaystyle \chi \displaystyle \sigma 12. Originally Posted by luobo Moo, I indeed is talking about series. For clarity, it is rearranged below: (1) Series \displaystyle \{b_n\} is convergent. \displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n { Note: \displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right) } The first term converges knowing \displaystyle \int^{\infty}_{1}\frac{\sin x}{x}\,dx is bounded, the second term converges even more obviously. (2) Series \displaystyle \{c_n\} is also convergent. \displaystyle \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n It converges since \displaystyle \int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx is bounded. \displaystyle \left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|$$\displaystyle \leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty$

(3) Series $\displaystyle \{a_n\}$ is convergent
$\displaystyle \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n$
you should know that the convergence of$\displaystyle \int^{\infty} f(x) \ dx$ doesn't necessarily imply that $\displaystyle \sum^{\infty} f(n)$ is convergent. for example $\displaystyle \int_0^{\infty} \sin x^2 \ dx=\sqrt{\frac{\pi}{8}}$ but the series $\displaystyle \sum_{n=0}^{\infty} \sin n^2$ is divergent.

13. NonConAlg,

You are right. Thanks a lot. Maybe we can switch to Abel's test. Abel's Test. Check below.

(1) Partial sum $\displaystyle \sum_{n=1}^{N}\sin n$ is bounded.
$\displaystyle \sum_{n=1}^{N}\sin n$ = Im $\displaystyle e^i+e^{2i}+e^{3i}+...+e^{Ni}$ = Im $\displaystyle \frac{e^i-e^{(N+1)i}}{1-e^i}$
$\displaystyle \left |\sum_{n=1}^{N}\sin n\right | \leq \left|\frac{e^i-e^{(N+1)i}}{1-e^i}\right |=\frac{\left|e^i-e^{(N+1)i}\right|}{\left|1-e^i\right|}$$\displaystyle \leq \frac{\left|e^i\right|+\left|e^{(N+1)i}\right|}{\l eft|1-e^i\right|}=\frac{2}{\left|1-e^i\right|}=\csc{\frac{1}{2}}$

(2) Series $\displaystyle \{b_n\}$ is convergent.
$\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n$ { Note: $\displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right)$ }
The second term converges obviously. The first term also converges according to Abel's test:
(Condition 1) Sequence $\displaystyle \left\{\frac{1}{n}\right\}$ is decreasing
(Condition 2) $\displaystyle \lim_{n\to\infty}\frac{1}{n}=0$
(Condition 3) Partial sum $\displaystyle \sum_{n=1}^{N}\sin n$ is bounded

(3) Series $\displaystyle \{c_n\}$ is also convergent.
$\displaystyle \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n$
(Condition 1) Sequence $\displaystyle \left\{\frac{\ln n}{n}\right\}$ is decreasing for $\displaystyle n=3,4,5...$
(Condition 2) $\displaystyle \lim_{n\to\infty}\frac{\ln n}{n}=0$
(Condition 3) Partial sum $\displaystyle \sum_{n=1}^{N}\sin n$ is bounded

(4) Series $\displaystyle \{a_n\}$ is convergent
$\displaystyle \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n$

14. The sequence $\displaystyle a_n=\frac{H_n}n$ is monotonically decreasing to $\displaystyle 0$.

Let $\displaystyle b_k=\sum_{n=1}^k \sin n$. Then $\displaystyle b_k\sin{\textstyle\frac12}=\sum_{n=1}^k \frac{\cos(n-{\textstyle\frac12})-\cos(n+{\textstyle\frac12})}2=\frac{\cos({\textsty le\frac12})-\cos(k+{\textstyle\frac12})}2$.

Thus $\displaystyle |b_k|=\left|\sum_{n=1}^k \sin n\right|\leq\frac1{\sin{\textstyle\frac12}}$ for all $\displaystyle k\in\mathbb N$, i.e. the partial sums of $\displaystyle \sum_{n=1}^\infty \sin n$ are bounded.

By Dirichlet's test, $\displaystyle \sum_{n=1}^\infty a_n\sin n$ converges.

15. halbard,

I was thinking about doing the same thing . My reference about the Abel's Test from a website appears ought to be the Dirichlet's test.

Luobo

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