Suppose $f:[0,1]\rightarrow\mathbb{R}$. defined by $f(x)=(-1)^n n $ when $x\in(1/(n+1),1/n]$ and $f(0)=0$. Show that the improper Riemann integral $$\int_{0}^{1} f(x) dx $$ is real number.
First note that $f$ has infinite points when $f$ is discontinuous. I have trouble to compute the integral. I have solved many problems but all of them was with finite points. Any help will be appreciated.
On
Let $\epsilon>0$ and let $n_0(\epsilon)\in\mathbb{N}$such that $\frac{1}{n_0+1}<\epsilon<\frac{1}{n_0}$:Notice $n_0(\epsilon)\underset{\epsilon\rightarrow0}{\rightarrow}\infty$ and $\frac{n_0(\epsilon)}{\epsilon}\underset{\epsilon\rightarrow0}{\rightarrow}1$
Now, $$\underset{\epsilon\rightarrow0}{lim} \int_{\epsilon}^{1}f(x)dx=\underset{\epsilon\rightarrow0}{lim}( \int_{1/n_0}^{1}f(x)dx+\int_{\epsilon}^{1/n_0}f(x)dx)=\\=\underset{\epsilon\rightarrow0}{lim}( \sum_1^{n_0(\epsilon)-1}(-1)^n\cdot n\cdot(\frac{1}{n}-\frac{1}{n+1})+\int_{\epsilon}^{1/n_0(\epsilon)}f(x)dx)=\\=\underset{\epsilon\rightarrow0}{lim}( \sum_1^{n_0(\epsilon)-1}(-1)^n\cdot (\frac{1}{n+1})+\int_{\epsilon}^{1/n_0(\epsilon)}f(x)dx)=0\\$$
Since- $$\underset{\epsilon\rightarrow0}{lim} \sum_1^{n_0(\epsilon)-1}(-1)^n\cdot (\frac{1}{n+1})=0$$ $$\underset{\epsilon\rightarrow0}{lim}\vert\int_{\epsilon}^{1/n_0(\epsilon)}f(x)dx\vert\leq(1-n_0\epsilon)\rightarrow0$$
Hint
Look at $$\lim_{n\to \infty }\sum_{k=1}^n\int_{1/(k+1)}^{1/k}(-1)^n n\mathrm d x.$$