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.
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 $$ |\gamma_n| = 1^n \cdot n \cdot \frac{1}{n(n+1)} \text{.} $$