The sequence of $$ a_x ={\cos (x)\over x} $$ does converge to zero. As a result, intuitively $$ \sum_{x=1}^\infty {\cos (x)\over x} $$ should also converge right? But I've been told that the series diverges. This shouldn't be true... right?

  • 2
    $\begingroup$ What is $n$ in your definition of $a_n$ ? Do you mean $cos(n)/n$? $\endgroup$
    – Horace
    Aug 19, 2018 at 11:40
  • 1
    $\begingroup$ Are you sure you formatted the question right? It isn't making much sense to me at the moment. $\endgroup$
    – prog_SAHIL
    Aug 19, 2018 at 11:41
  • $\begingroup$ sorry, i'll edit it $\endgroup$ Aug 19, 2018 at 11:41
  • $\begingroup$ When $n=0,$ the term $\cos(0)/0$ is not defined. Do you mean to sum from $n=1$ to $\infty$? $\endgroup$
    – coffeemath
    Aug 19, 2018 at 11:46
  • 1
    $\begingroup$ If a series converges, then the limit of the general term must be $0$. The converse is not true, as the usual example of the harmonic series shows. $\endgroup$
    – MasB
    Aug 19, 2018 at 11:54

1 Answer 1


It is not at all intuitive to me that the series ought to converge simply because the terms go to zero. For example $\sum_{n=1}^{\infty} \frac1n$ is well known to diverge even though $\frac1n\to 0$. Or, as an even easier example, consider $$ 1 + \underbrace{\frac12+ \frac12}_{2\text{ halves}} + \underbrace{\frac13 + \frac13+ \frac13}_{3\text{ thirds}} + \underbrace{\frac14 + \frac14 + \frac14+ \frac14}_{4\text{ fourths}}+ \underbrace{\frac15 + \frac15+ \frac15+ \frac15+\frac15}_{5\text{ fifths}}+ \cdots $$

It does look like your particular series converges (conditionally), by Dirichlet's test, though.


Not the answer you're looking for? Browse other questions tagged .