here's my question:
Suppose $f:\mathbb{R}_+\to\mathbb{R}_+$ continuous and increasing.
Prove that: $\sum f(e^{-n})$ and $\sum \dfrac 1 n f\left(\dfrac 1 n \right)$ are both convergent or divergent.
p.s. A tip says I could try to use integration, but I don't know how...
Thans a lot~
The key is to apply the Integral Test for convergence of series to both of them. Interestingly enough, after substitutions both integrals turn into the same one!
For the first series $\sum f(e^{-n})$, consider the function $h(x)=f(e^{-x})$ on $x\in[0,+\infty)$. Because of the given conditions on $f$, $h$ is positive, continuous, and decreasing, so the Integral Test applies, and then given series converges or diverges together with the improper integral $\int_0^{+\infty}h(x)\,dx=\int_0^{+\infty}f(e^{-x})\,dx$. Now let's substitute $t=e^{-x}$; then $x=-\ln t$ and $dx=-\frac{1}{t}dt$. So the integral becomes $$\int_0^{+\infty} h(x)\,dx=\int_0^{+\infty} f(e^{-x})\,dx=\int_1^0 f(t)\cdot\frac{-1}{t}\,dt=\int_0^1 \frac{1}{t}f(t)\,dt,$$ improper at $t=0$.
For the second series $\sum \frac{1}{n}f(\frac{1}{n})$, consider the function $g(x)=f(\frac{1}{n})$ on $x\in(0,+\infty)$. Because of the given conditions on $f$, $g$ is positive, continuous, and decreasing, so the Integral Test applies, and then given series converges or diverges together with the improper integral $\int_0^{+\infty}g(x)\,dx=\int_0^{+\infty}\frac{1}{x}f(\frac{1}{x})\,dx$. Now let's substitute $t=\frac{1}{x}$; then $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$. So the integral becomes $$\int_0^{+\infty} g(x)\,dx=\int_0^{+\infty} \frac{1}{x}f\left(\frac{1}{n}\right)\,dx=\int_1^0 tf(t)\cdot\frac{-1}{t^2}\,dt=\int_0^1 \frac{1}{t}f(t)\,dt,$$ improper at $t=0$.
So both series converge or diverge together with the same improper integral.