Infinite series (4)

**NonCommAlg** · May 22nd 2009, 09:28 PM

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

**simplependulum** · May 23rd 2009, 09:47 PM

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

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

**NonCommAlg** · May 23rd 2009, 10:16 PM

Originally Posted by simplependulum

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

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

**Sampras** · May 24th 2009, 05:47 AM

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

**chisigma** · May 24th 2009, 11:47 PM

If we indicate with...

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

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

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

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

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

... we derive in two steps...

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

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

Now we use the identity...

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

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

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

Kind regards

$\chi$ $\sigma$

**luobo** · August 6th 2009, 09:26 PM

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

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

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

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

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

**Moo** · August 6th 2009, 11:43 PM

Luobo,

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

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

**luobo** · August 7th 2009, 03:46 AM

Moo,

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

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

(2) Series $\{c_n\}$ is also convergent.
$\sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n$
It converges since $\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx$ is bounded.
$\left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|$ $\leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty$

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

**chisigma** · August 7th 2009, 08:33 PM

Originally Posted by Moo

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

... all right!... with little patience [or even using Wolfram Integrator

...] you find that...

$\varphi (x)= - \int_{0}^{x} \frac{\ln(1-\xi)}{\xi(1-\xi)} d\xi= \frac{1}{2} \ln^2 (1-x) + Li_{2} (x)$ (1)

... where $Li_{2} (*)$ is the Polylogarithm function of order two, defined in the complex plane as...

$Li_{2} (z)= \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}}$ (2)

Now the series (2) converges for $|z|=1$ so that $\varphi (e^{i})$ and $\varphi (e^{-i})$ both exist...

Kind regards

$\chi$ $\sigma$

**chisigma** · August 8th 2009, 10:05 PM

Following the same procedure we arrive to demonstrate that is...

$\sum_{n=1}^{\infty} H_{n} \frac{\cos n}{n} = \frac{1}{2} \{\varphi (e^{i}) + \varphi(e^{-i})\}$ (1)

These are interesting results that are not isolated, because the same procedure can be used to arrive to even better results

...

Kind regards

$\chi$ $\sigma$

**chisigma** · August 8th 2009, 11:40 PM

An example... from the well known expression...

$\sum_{n=1}^{\infty} x^{n-1}= \frac{1}{1-x}$ (1)

... we derive...

$\sum_{n=1}^{\infty} \frac{x^{n}}{n} = \int_{0}^{x} \frac{d\xi}{1-\xi} = -\ln (1-x)$ (2)

Now from (2) and the identity...

$\sin n= \frac{e^{in}-e^{-in}}{2i}$ (3)

... we derive...

$\sum_{n=1}^{\infty} \frac{\sin n}{n} = \frac{\ln (1-e^{-i}) - \ln (1-e^{i})}{2i}= \tan^{-1} \frac {\sin 1}{1-\cos 1} = 1.07079632679\dots$ (4)

... and from (2) and the identity...

$\cos n= \frac{e^{in}+e^{-in}}{2}$ (5)

... we derive...

$\sum_{n=1}^{\infty} \frac{\cos n}{n} = \frac{\ln (1-e^{-i}) + \ln (1-e^{i})}{2}= \ln \sqrt {2 (1-\cos 1)} = -.042019505825\dots$ (6)

Kind regards

$\chi$ $\sigma$

**NonCommAlg** · August 9th 2009, 09:42 AM

Originally Posted by luobo

Moo,

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

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

(2) Series $\{c_n\}$ is also convergent.
$\sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n$
It converges since $\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx$ is bounded.
$\left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|$ $\leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty$

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

you should know that the convergence of $\int^{\infty} f(x) \ dx$ doesn't necessarily imply that $\sum^{\infty} f(n)$ is convergent. for example $\int_0^{\infty} \sin x^2 \ dx=\sqrt{\frac{\pi}{8}}$ but the series $\sum_{n=0}^{\infty} \sin n^2$ is divergent.

**luobo** · August 9th 2009, 04:05 PM

NonConAlg,

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

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

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

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

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

**halbard** · August 9th 2009, 04:49 PM

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

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

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

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

**luobo** · August 9th 2009, 05:28 PM

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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