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So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:

Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.

I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.

Sujaan Kunalan
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user87479
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    Start at $0$. Add small numbers till you get to $1$. Subtract smaller numbers till you get to $-1$. Add even smaller numbers till you get to $1$ again. ... – David Mitra Jul 24 '13 at 00:16
  • @AlexMardikian I think David meant the "small numbers" to be the sequence $a_n$, and the sum (which oscillates) to be $s_n$. – Pedro M. Jul 24 '13 at 00:21
  • @PedroMilet Thanks, I read that the wrong way. – Alex Jul 24 '13 at 00:24
  • Of course! It's just a slight modification of the simple harmonic series. – user87479 Jul 24 '13 at 00:41
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    David has given a perfect hint/answer. Note that by Monotone Convergence Theorem, if the sequence ${s_n}_{n=1}^{\infty}$ is bounded and monotonically increasing (or decreasing), then it must converge. So any possible example will have to involve some sort of oscillation. – Prism Jul 24 '13 at 00:45

2 Answers2

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For the sake of having an answer:

$S_n$ need not converge. To see this, here is a

Hint:

Start at $0$. Add small numbers till you get to $1$. Subtract smaller numbers till you get to $−1$. Add even smaller numbers till you get to $1$ again. ...


Perhaps this is a full solution; but I could think of no way to phrase a "hint" that is both useful and not a full solution. Perhaps Prism's comment would be enough.

David Mitra
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Try showing the sequence $a_n = \sin(\ln(n)) - \sin(\ln(n-1))$ is a conterexample.

Zarrax
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