0

I have this integral $$ \int_{0}^{1}f(t)dt $$ where $$f(t)=(-1)^n \cdot n$$ for $$\frac{1}{n+1}< t\leq \frac{1}{n}, n\epsilon \mathbb{N}$$

I have to show that is converges but does not converge absolutely. I started like this $$D=\left \{ t:0< t\leq 1 \right \} $$ $$D_{n}=\left \{ t:\frac{1}{n+1}< t\leq \frac{1 }{n}\right \}$$ $$ \gamma_{n}=\int_{\frac{1}{n+1}}^{\frac{1}{n}}f(t)dt=\int_{\frac{1}{n+1}}^{\frac{1}{n}}(-1)^n \cdot ndt$$ And then I wanted to calculate $$\gamma_{n} n \to \infty$$ but something is not right. I could use some help.

Awerde
  • 299
  • Why do you say "something is not right"? What were you expecting to happen that did not? – Eric Towers May 25 '20 at 15:25
  • Well, the limit exists, though it is very strange and I don't know if it should be like that, but what is not right for certain is that the absolute convergance does happen and I have a feeling I don't understand they way it should be calculated. I think that when we take absolute value $$\left | f(t) \right |=\left | n \right |$$ but that doesn't work. – Awerde May 25 '20 at 15:35
  • $|\gamma_k| = 1 \cdot n \cdot \frac{n}{n(n+1)} = \frac{1}{n+1}$ does not lead to absolute convergence. You should perhaps show the work that is leading you to confusion. – Eric Towers May 25 '20 at 15:42
  • But isn't $\left | \gamma_{n} \right |=\frac{1}{n+1} n \to \infty = 0$? Am I missing something? – Awerde May 25 '20 at 15:51

1 Answers1

0

I iterate: "You should perhaps show the work that is leading you to confusion."

For instance, $$ \frac{1}{n} - \frac{1}{n+1} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)} \text{,} $$ so $$ |\gamma_n| = 1^n \cdot n \cdot \frac{1}{n(n+1)} \text{.} $$

Eric Towers
  • 67,037
  • Okay, but, as I commented before, $\lim_{n \to \infty}\left | \gamma_{n} \right |=\lim_{n \to \infty}\frac{1}{n+1} = 0$ so the limits exists thus integral has absolute convergence and I had to show it doesn't so I suppose I don't see something. – Awerde May 25 '20 at 16:18
  • @Awerde : $\sum_n \frac{1}{n+1}$ is not finite. This sum does not converge. What convergence test do you think you are applying? – Eric Towers May 25 '20 at 16:29
  • Oh, I see now. Thank you. – Awerde May 25 '20 at 16:40