In my Analysis class the other day we were discussing Sequences that have bounded partial sums yet their infinite series does not converge, with the typical example of $\{ a_n \}=(-1)^n$. In discussion we started to wonder if there exists an unbounded sequence that had bounded partial sums.

I have been thinking about this for a few days and I think that $$b_n = (-1)^nln(n) $$ is such a sequence. In Mathematica I tested the first 2,000,000 sums and they are all within $\pm10 $. I would like to be able to prove that the partial sums are bounded yet I really do not know even where to start. If you can come up with any other examples or know where to start proving $$ \sum_1^n (-1)^nln(n) $$ is bounded, I would love to hear what you have to say.


1 Answer 1


No you cannot have something like that.

Assume that your partial sums are bounded between $\pm M, M>0$

Then since your sequence is unbounded(WLOG from above) there would be an $a_{n_0} >3M$

Then if you take $$ \sum _{i=1}^{n_0}a_i = \sum _{i=1}^{n_0-1}a_i+a_{n_0} >-M +3M > 2M$$ a contradiction

  • $\begingroup$ Thanks. I'm realizing I was was under the assumption it had to exist. I blinded myself. Again thanks. $\endgroup$
    – Garrett N
    Jan 17, 2014 at 7:02
  • $\begingroup$ No problem, yes that happens! $\endgroup$
    – clark
    Jan 17, 2014 at 7:05

You must log in to answer this question.

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