This discussion was brought up in my real analysis course today and it was left mostly unresolved so I was looking for clarification.
Basically, a sequence that does not converge can either properly diverge (be unbounded) or have multiple subsequences with different limits. That is well and good.
A series however, can be said to converge as long as it's less than another convergent sequence (Comparison Test). This seems fine too until you consider a series as a sequence of partial sums. What specifically prevents this sequence of partial sums from being neither properly divergent nor convergent, i.e., having multiple subsequences that converge to two distinct limits, both of which are less than another convergent series?

  • $\begingroup$ To apply the comparison test, it must be in relation to a series with all non-negative terms. As for why it must converge, it follows from the monotone convergence theorem. $\endgroup$
    – JMoravitz
    Oct 19, 2016 at 21:26
  • $\begingroup$ ah, being strictly positive is a good point which wasn't made clear in class thank you! $\endgroup$ Oct 19, 2016 at 21:32
  • $\begingroup$ Terms which are zero are fine. The point is that the sequence of partial sums should be a (nonstrictly) increasing sequence bounded above by something (what the "larger" series converged to). If you remove the non-negative requirement, you could have things like $\sum\limits_{i=1}^\infty -1 \leq \sum\limits_{i=1}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}$, and although $-1\leq \frac{1}{i^2}$ for every $i$, the summation on the far left clearly diverges. You can still get around it by considering the absolute value of the series. $\endgroup$
    – JMoravitz
    Oct 19, 2016 at 21:35

1 Answer 1


This is merely due to a missed requirement to apply the comparison test. To apply the comparison test, you must have all terms strictly positive to set the following bounds:


Or else $\sum_{n\ge1}a_n$ would be unbounded in the negative direction.

The condition that $a_n\ge0$ also allows the Lebesgue monotone convergence theorem that a bounded increasing sequence converges to come into play, which forces $\sum_{n\ge1}a_n$ to converge to a single value if it converges.

  • $\begingroup$ the monotone convergence theorem ? No, you mean the theorem that a bounded increasing sequence converges (and the coefficients $a_n$ have to be positive (non-negative), not strictly positive) $\endgroup$
    – reuns
    Oct 19, 2016 at 23:15
  • $\begingroup$ @user1952009 the monotone convergence theorem is the theorem that says a bounded monotonic sequence (e.g. increasing) must converge. Next, the phrase "positive" and "strictly positive" mean the same thing. Zero is not a positive number, nor is it a strictly positive number. Non-negative does not mean the same thing as positive, as zero is a non-negative number. $\endgroup$
    – JMoravitz
    Oct 20, 2016 at 1:09
  • $\begingroup$ @JMoravitz : Well in french the "theoreme de convergence monotone" is only about inverting limit and $\int$ (or $\lim$ and $\sum$ or $\sum$ and $\int$), i.e. it is the Lebesgue monotone convergence theorem $\endgroup$
    – reuns
    Oct 20, 2016 at 1:16
  • $\begingroup$ @JMoravitz See google I'm not dreaming, "monotone convergence theorem" for almost everybody refers to the Lebesgue MCT (also called the Beppo-Levi theorem) not the theorem that a bounded monotone sequence converges $\endgroup$
    – reuns
    Oct 20, 2016 at 1:25
  • $\begingroup$ @user1952009 Ok... sure I'll change that. $\endgroup$ Oct 20, 2016 at 14:07

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