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This is a counter example homework question that I can't seem to solve.

I need to find a sequence of real numbers $(x_n)_{n=1}^{\infty}$, and a monotonic increasing sequence of natural numbers $(n_k)_{k=1}^{\infty}$such that: $(i)\lim\limits_{n\to\infty}(x_n-x_{n+1})=0$, $(ii)(x_{n_k})_{k=1}^{\infty}$ converges, but $(iii)(x_n)_{n=1}^{\infty}$ do not converges.

Only thing I know so far is that for all $k$, $[n_{k+1}-n_k]$ cannot be bounded.

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it converges to 0. –  Amihai Zivan Jun 6 '12 at 14:43
I changed the term "sequence" to "series". Pardon my English. –  Amihai Zivan Jun 6 '12 at 14:45
@AmihaiZivan: I think "sequence" is what you meant. The word "series" implies that you're summing the terms, but that is not what your formulas say -- they speak about the raw elements of the sequence converging. –  Henning Makholm Jun 6 '12 at 14:47
I know this is calculus, but I found this to be an enlightening book on series and sequences. amazon.com/Counterexamples-Analysis-Dover-Books-Mathematics/dp/… –  rckrd Jun 6 '12 at 14:54
@Henning Thanks. I'll change that once again.. :-) –  Amihai Zivan Jun 6 '12 at 15:10
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2 Answers 2

up vote 3 down vote accepted

It's sometimes easiest to go for the most blatant counterexample you can get away with. So decide that you will have $x_{n_k}=0$ for all $k$ and $x_n=1$ for some $n$ between each set of $x_{n_k}$ and $x_{n_{k+1}}$.

This calls for a sequence that climbs from 0 to 1 and back down to 0 and up again infinitely many times. It must eventually do this slower and slower due to the condition on the successive difference, but you can decide how slow or fast the differences go to 0.

For example, decide that $|x_n-x_{n+1}|$ must be $1/k$ whenever $n$ is between $n_k$ and $n_{k+1}$...

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Hint: consider the sequence $0\,,{1\over2}\,,1\,,{2\over3}\,,{1\over3}\,,0,\,{1\over4}\,,{2\over4}\,,{3\over4}\,,1\,,{4\over5}\,,{3\over5}\,,{2\over5}\,,{1\over5}\,,0\,,{1\over6}\,,\ldots$ .

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+1 This is a really neat example. –  Derek Allums Jun 6 '12 at 14:49
Excellent! Thanks David Mitra. –  Amihai Zivan Jun 6 '12 at 15:13
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