0

I'm trying to establish if the following statement is true.

If $ a_{n}$ is a sequence of real postive numbers such that $ \sum_{n=0}^{\infty} a_{n} <\infty, $ than $ a_{n} = o\left(\frac{1}{n}\right). $

If it is not true, can anyone show me a counterexample?

1 Answers1

5

Consider the sequence given by $a_n=n^{-1}$ if $n$ is a perfect square and $a_n=0$ otherwise. Then it is easy to see that $\sum a_n<\infty$ but $na_n\not\to 0$

Sri-Amirthan Theivendran
  • 31,832
  • Could you explain me why $ \sum a_{n} < \infty? $ i was thinking about $ \sum \frac{1}{n^{3}}. $ It is convergent and $n\cdot(n^3)\nrightarrow 0. $ It is wrong? –  Jun 01 '19 at 16:25
  • 2
    You should define $a_n =1/2^n$ or something for $n$ not a perfect square. – zhw. Jun 01 '19 at 16:41
  • My bad thought positive meant non-negative. If you need strictly positive go with zhw's comment – Sri-Amirthan Theivendran Jun 01 '19 at 17:57