Convergence and absolute convergence of alternating series $\sum\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t}dt$

Question

I have to determine the convergence (later absolute convergence) of the alternating series with the general term

$$v_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t}dt$$

using the usual alternating series rule: $\sum (-1)^n u_n$ converges if $(u_n)$ is positive and decreases to $0$.

So, if $n$ is even, $v_{n}$ is positive due to positivity of $\sin t$ in the interval $[n\pi,(n+1)\pi]$ and if $n$ is odd then $v_n$ is negative by negativity of $\sin t$ in the corresponding interval. Then I can remark that

$$v_n=(-1)^n\int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t}dt=(-1)^n|v_n|$$

and since $|v_n|$ is lower bounded by $0$, by monotone convergence theorem, we just have to prove that $|v_n|$ is decreasing. Here, I can say that since $\frac{1}{t}$ decreases and $|\sin t|$ is $\pi$-periodic, then $|v_n|$ decreases. Thus $\sum v_n$ converges.

By symmetry of $n$, $|v_n|=(-1)^{-n}v_n=(-1)^{n}v_n$. We write (let's say $n$ is odd)

$$\sum (-1)^{n}v_n = v_0 - v_1 + v_2 - v_3 + \dots + v_{n-1} - v_n$$

So $\sum|v_n|=\sum(v_{n-1} - v_n)$ is a telescoping series and since $\sum v_n$ converges then by linearity of convergent sums $\sum|v_n|$ also converges.

But is there a better way to solve this problem? I feel like my reasoning is flawed here, especially when proving that $|v_n|$ decreases.

Answer 1

You have to prove that $v_n\rightarrow 0$, the fact that $(|v_n|)$ is lower bounded by $0$ and is decreasing doesn't mean that $v_n\rightarrow 0$, take $v_n=1+\frac{1}{n}$ for instance. But, $$ |v_n|=\int_{n\pi}^{(n+1)\pi}\frac{|\sin(t)|}{t}dt\leqslant\int_{n\pi}^{(n+1)\pi}\frac{dt}{t}=\log\left(1+\frac{1}{n}\right)\underset{n\rightarrow +\infty}{\longrightarrow} 0 $$ Here is another proof : $$ \sum_{n=1}^{N-1} v_n=\int_0^{N\pi}\frac{\sin(t)}{t}dt $$ An integration by parts gives that $$ \sum_{n=1}^{N-1}v_n=\frac{1-(-1)^N}{N\pi}+\int_0^{N\pi}\frac{1-\cos(t)}{t^2}dt $$ Therefore, it suffices to show that $\displaystyle\lim\limits_{N\rightarrow +\infty}\int_0^{N\pi}\frac{1-\cos(t)}{t^2}dt$ exists and is finite. It is indeed the case because this is a non-decreasing sequence, bounded above because $$ \int_0^{N\pi}\frac{1-\cos(t)}{t^2}dt\leqslant\int_0^1\frac{1-\cos(t)}{t^2}dt+\int_1^{N\pi}\frac{2}{t^2}dt\leqslant\int_0^1\frac{1-\cos(t)}{t^2}dt+2 $$

NOTE : You also conclude that since $\sum v_n$ converges, then $\sum |v_n|$ also converges, this is absolutely false, take $v_n=\frac{(-1)^n}{n}$ for instance. This does not hold because you can't always change the order of summation when $v_n$ is not of constant sign (though you can when $\sum |v_n|$ converges). What you just did is this : $$ (v_0-v_1)+(v_1-v_2)+\ldots=v_0+(-v_1+v_1)+(-v_2+v_2)+\ldots=v_0 $$ which you can't do because of what said above !

Answer 2

$v_n =\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t}dt =(-1)^n\int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t}dt $

$\begin{array}\\ v_n+v_{n+1} &=\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t}dt+\int_{(n+1)\pi}^{(n+2)\pi} \frac{\sin t}{t}dt\\ &=\int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t}dt+\int_{n\pi}^{(n+1)\pi} \frac{\sin (t+\pi)}{t+\pi}dt\\ &=\int_{n\pi}^{(n+1)\pi} \left(\frac{\sin t}{t}+\frac{\sin (t+\pi)}{t+\pi}\right)dt\\ &=\int_{n\pi}^{(n+1)\pi} \left(\frac{\sin t}{t}-\frac{\sin (t)}{t+\pi}\right)dt\\ &=\int_{n\pi}^{(n+1)\pi} \sin(t)\left(\frac1{t}-\frac1{t+\pi}\right)dt\\ &=\int_{n\pi}^{(n+1)\pi} \sin(t)\frac{\pi}{t(t+\pi)}dt\\ &=\pi\int_{n\pi}^{(n+1)\pi} \frac{\sin(t)}{t(t+\pi)}dt\\ v_{2n}+v_{2n+1} &=\pi\int_{2n\pi}^{(2n+1)\pi} \frac{\sin(t)}{t(t+\pi)}dt\\ &=\pi\int_{0}^{\pi} \frac{\sin(t+2n\pi)}{(t+2n)(t+(2n+1)\pi)}dt\\ &=\pi\int_{0}^{\pi} \frac{\sin(t)}{(t+2n)(t+(2n+1)\pi)}dt\\ &\gt 0\\ \text{and}\\ v_{2n}+v_{2n+1} &\lt\pi\int_{0}^{\pi} \frac{1}{(t+2n)(t+(2n+1)\pi)}dt\\ &\lt\pi\int_{0}^{\pi} \frac{1}{(t+2n)^2}dt\\ &\lt \dfrac{\pi^2}{4n^2}\\ \end{array} $

and the sum of these converges.