Problem. Discuss the convergence of the following two series, $$\displaystyle\sum_{i=3}^\infty \left( 1-\dfrac{\ln i}{i}-\left(\dfrac{\ln(\ln i)}{i}\right)\left(\cos^2\dfrac{1}{i}\right)(a+(-1)^ib)\right)^i$$$$\displaystyle\sum_{i=3}^\infty \left( 1-\dfrac{\ln i}{i}-\dfrac{c\ln(\ln i)}{i}\right)^i$$where $a,b,c\in\mathbb{R}$

So far I have observed that, $$\left( 1-\dfrac{\ln i}{i}-\left(\dfrac{\ln(\ln i)}{i}\right)\left(\cos^2\dfrac{1}{i}\right)(a+(-1)^ib)\right)^i\le \left( 1-\dfrac{\ln i}{i}-\dfrac{(|a|+|b|)\ln(\ln i)}{i}\right)^i$$So, if we can show that the second series converges for all $c\in\mathbb{R}^+\{0\}$ then we can claim that the first series also converges.

But I don't know how to proceed from here. Can anyone help?


1 Answer 1



$$\begin{align} \left(1-\frac{f(i)}{i}\right)^i&=e^{i\log\left(1-\frac{f(i)}{i}\right)}\\\\ \end{align}$$


$$-\frac{f(i)}{i-1}\le\log\left(1-\frac{f(i)}{i}\right)\le -\frac{f(i)}{i}$$

where here $f(i)=\log(i)+c\log(\log(i))$

  • $\begingroup$ I am sorry that I am late. But actually I was trying to work out your hint but unfortunately couldn't see how to solve the problem. Can you tell me what I should do from here? $\endgroup$
    – user170039
    Apr 2, 2016 at 12:52
  • $\begingroup$ Use the comparison test. $\endgroup$
    – Mark Viola
    Apr 2, 2016 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy