Prove that the series is non-absolutely convergent.

Question

$$a_n = \int_{(2n-2)\pi}^{(2n-1)\pi} \dfrac{\sin t}{t} dt$$

The series is $$\sum_{n=1}^{\infty}a_n$$

I tried using the Cauchy criterion, and this let me with the next inequality:

$$\left| S_m - S_n \right| \le \left|\int_{(2(n+1)-2)\pi}^{(2(n+1)-1)\pi} \dfrac{\sin t}{t}dt\right| + \cdots +\left|\int_{(2m-2)\pi}^{(2m-1)\pi}\dfrac{\sin t}{t}dt\right| \le \int_{2n\pi}^{(2n+1)\pi}\dfrac{1}{t}dt+\cdots+\int_{(2m-2)\pi}^{(2m-1)\pi}\frac 1t dt$$

But I don't know if I need evaluete the integrals or make another inequality with them and then integrate.

Possible doublicate. – Mhenni Benghorbal Apr 29 '13 at 05:42 — Mhenni Benghorbal, Apr 29 '13 at 05:42

score 1 · Answer 1 · answered Apr 29 '13 at 05:57

1

Here's a big hint.

$$|a_n| = \int_{(2n-2)\pi}^{(2n-1)\pi}\frac{|\sin t|}{t}\,dt > \int_{(2n-\frac{11}{6})\pi}^{(2n-\frac{7}{6})\pi}\frac{|\sin t|}{t}\,dt> \int_{(2n-\frac{11}{6})\pi}^{(2n-\frac{7}{6})\pi}\frac{1/2}{t}\,dt{\textrm.}$$

answered Apr 29 '13 at 05:57

Steve Kass

14,881

score 1 · Accepted Answer · answered Apr 29 '13 at 05:57

1

Since $\sin{t}\geqslant{0}$ on $[(2n-2)\pi,\ (2n-1)\pi]$ and $\sin{t}\geqslant\dfrac{1}{2}$ for $t\in \left[(2n-2)\pi+\frac{\pi}{6},\; (2n-1)\pi-\frac{\pi}{6}\right]$ $$a_n = \int\limits_{(2n-2)\pi}^{(2n-1)\pi} \dfrac{\sin t}{t} dt \geqslant \int\limits_{(2n-2)\pi+\frac{\pi}{6}}^{(2n-1)\pi-\frac{\pi}{6}} \dfrac{\sin t}{t} dt \geqslant \dfrac{1}{2}\int\limits_{(2n-2)\pi+\frac{\pi}{6}}^{(2n-1)\pi-\frac{\pi}{6}} \dfrac{dt}{t} =\\ =\dfrac{1}{2}\ln{\dfrac{(2n-1)\pi-\frac{\pi}{6}}{(2n-2)\pi+\frac{\pi}{6}}}=\dfrac{1}{2}\ln{\dfrac{12n-7}{12n-11}}=\dfrac{1}{2}\ln\left(1+\dfrac{4}{12n-11} \right)\underset{n\to\infty}\sim\dfrac{2}{12n-11}>\dfrac{1}{6n}$$

answered Apr 29 '13 at 05:57

M. Strochyk

8,397

There's no need to integrate (not that it's hard). $\int_a^b \frac{dt}{t} > \int_a^b \frac{dt}{b}= \frac{b-a}{b}$, which with the values of $a$ and $b$ you have will take care of things. – Steve Kass Apr 29 '13 at 06:02
But, the last limit goes to 0? – David Apr 29 '13 at 06:13
1

Since the harmonic series $ \sum\limits_{n=1}^{\infty}{\frac{1}{n}}$ diverges and ${a_n} > \frac{1}{6}\cdot{\frac{1}{n}}$ then $\sum\limits_{n=1}^{\infty}{a_n}$ is divergent by comparison test. – M. Strochyk Apr 29 '13 at 06:46
Of course. But the step between the log and $\dfrac{2}{12n-11}$ is not clear for me. – David Apr 29 '13 at 06:52
I used the equivalence $\ln(1+x)\underset{x\to{0}}\sim{x}$ with $x=\dfrac{4}{12n-11}.$ – M. Strochyk Apr 29 '13 at 07:01
Were I can find more info about that @M.Strochyk? – David Apr 30 '13 at 00:32
The equivalence $\ln(1+x)\underset{x\to{0}}\sim{x}$ is another form of $\lim\limits_{x\to{0}}{\dfrac{\ln(1+x)}{x}}=1,$ which can be found in every calculus textbook. – M. Strochyk Apr 30 '13 at 06:48

Prove that the series is non-absolutely convergent.

2 Answers2