finiteness of $\sum_n a_n$ vs. $\sum_n \varphi(a_n)$

Question

There are some questions that ask for a (non)finiteness relation between $\sum_n a_n$ and $\sum_n \varphi(a_n)$:

• $\varphi(x)=\frac{x}{1+x}$ see: 1

• $\varphi(x)=\log(\frac{1}{1-x})$ see: 2

• $\varphi(x)=\sin(x)$ see: 3

Let me share a simple approach, a practical tool to handle some of these questions:

Simple version

Suppose that:

• $0\le a_n\le C$ for some $C>0$
• $\varphi$ is a real function
  • $\varphi(0)=0$
  • $0<L\le\varphi'(x) \le U$ on $[0,C]$ for some positive reals $L,U$.

Then: $$ \boxed{\sum_n a_n \ \text{ (in)finite} \iff \sum_n \varphi(a_n)\ \text{ (in)finite}} $$

"Proof":
By Lagrange's mean-value theorem: $$ \varphi(a_n)=\varphi(a_n)-\varphi(0)=\varphi'(\xi)a_n \implies\\ L\ a_n \le \varphi(a_n) \le U\ a_n $$

Using this proposition it is easy to solve the problems mentioned in the intro.

I am aware that the restrictions for $\varphi$ are too hard, for example it cannot handle the convergence of $\sum_n \frac{\log(1+\frac{1}{n})}{n}$, because the vanishing derivative of $\varphi(x)=x\log(1+x)$ at $0$. It can be seen that a similar proposition (see the "Advanced" version below) can be stated involving higher derivatives that can handle a more general class of sequences. But this one is simple enough to remember (I think) and encourages the solver to search after patterns instead of apply ad-hoc estimations.

Question:

• do you know a different but simple and general way to handle these kind problems?

"Advanced" version

Suppose that:

• $0\le a_n\le C$ for some $C>0$
• $\varphi$ is a real, $p\ge 1$-times continously differentiable function on an open interval that contains $[0,C]$
  • $\varphi(0)=\ldots =\varphi^{(p-1)}(0)=0$
  • $\varphi^{(p)}>0$ on $[0,C]$

Then: $$ \boxed{\sum_n a^p_n \ \text{ (in)finite} \iff \sum_n \varphi(a_n) \text{ (in)finite}} $$

"Proof":
Denote by $L$ and $U$ the extremas of the continous $\varphi^{p}$ on $[0,C]$. By the Taylor expansion of $\varphi$ around $0$: $$ \varphi(a_n)=\varphi(0+a_n)=\sum_{r=0}^{p-1} \frac{\varphi^{(r)}(0)}{r!}a_n^r+ \frac{\varphi^{(p)}(\xi)}{p!}a_n^p = \frac{\varphi^{(p)}(\xi)}{p!}a_n^p \implies \\ \frac{L}{p!}\ a^p_n \le \varphi(a_n) \le \frac{U}{p!}\ a^p_n $$

Examples:

• $\sum_n \frac{\log(1+\frac{1}{n^p})}{n^p}<\infty$ for $p>\frac{1}{2}$
  • $\varphi(x)=x\log(1+x),\ \varphi(0)=0$
  • $\varphi'(x)=\log(1+x)+\frac{x}{1+x},\ \varphi'(0)=0$
  • $\varphi''(x)=\frac{1}{1+x}+\frac{1}{1+x}-\frac{x}{(1+x)^2}=\frac{1}{1+x}+\frac{1}{(1+x)^2}>0$ on $[0,1]$
  • $\varphi''$ continous (for example) on $(-\frac{1}{2},\frac{3}{2})$
  • the finiteness of $\sum_n \frac{1}{n^{2p}}$ gives the claim.
• $\sum_n \frac{e^\frac{1}{n^p}-1}{n^p}<\infty$ for $p>\frac{1}{2}$
  • $\varphi(x)=x(e^x-1),\ \varphi(0)=0$
  • $\varphi'(x)=e^x-1+xe^x,\ \varphi'(0)=0$
  • $\varphi''(x)=e^x+e^x+xe^x=(x+2)e^x>0$ on $[0,1]$
  • $\varphi''$ continous (for example) on $(-\frac{1}{2},\frac{3}{2})$
  • the finiteness of $\sum_n \frac{1}{n^{2p}}$ gives the claim.

Answer 1

We suppose that $a_n \geq 0$.

If $(a_n)$ tends to $0$, then $$a_n \sim \frac{a_n}{1+a_n} \sim \ln \left( \frac{1}{1-a_n} \right) \sim \sin(a_n)$$

so the convergences of all the series are equivalent.

If $(a_n)$ does not tend to $0$, then $\frac{a_n}{1+a_n}$ and $\ln \left( \frac{1}{1-a_n} \right)$ don't tend to $0$, so the three series $\sum a_n$, $\sum \frac{a_n}{1+a_n}$ and $\sum \ln \left( \frac{1}{1-a_n} \right)$ cannot converge. However, the series $\sum \sin(a_n)$ can converge.