Common mistakes in concluding that series is convergent or not Assuming $(na_{n})_{n=1}^{\infty}$ convergent to $0$ then
$\sum_{n=1}^{\infty}a_{n}$ is convergent, true or false ?
wel' I say it is true because:
$n$ convergent to infinity and $na_{n}$ is convergent to zero
meaning is also convergent to zero.
So the question who is "convergent faster".
I can conclude that $a_{n}$ is convergent faster then $\frac{1}{n}$ to $0$
because $n\frac{1}{n}$ convergent to $1$.
meaning $\exists \alpha \in \mathbb{R}_{> 1}\ \exists N \in \mathbb{N}\ \forall n>N \ a_{n} \leq \frac{1}{n^{\alpha}}$
And we know that $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ is convergent there fore the tail of  $\sum_{n=1}^{\infty}a_{n}$ is convergent meaning the series is convergent.
I was told this is not true and the answer is false.
So can someone please give a counterexample and please explain to me what I did wrong in my proof.
Thanks in advanced !!

Just because $na_n \to 0$, it doesn't follow that $a_n < n^{-\alpha}$ for some $\alpha > 1$.
Counterexample: $a_n = \dfrac{1}{n\log n}$. The corresponding series diverges; the simplest way to see this is via the integral test (or Cauchy's condensation test if you know that one).
