Saturday, October 18, 2014
A Problem
Problem. Suppose that (i) $\{x_n\}$ is bounded and (ii) $\limn (x_{n+k}-x_n)=0$ for every $k\in \N$. Is $\{x_n\}$ convergent?
Solution.
After a moment of thought one easily deduces that a sequence converges if and only if $\limn (x_{n+k}-x_n)=0$ uniformly in $k$. Therefore if every such sequence converges, the ``pointwise'' convergence in $k$ and ``uniform pointwise" convergence in $k$ will be the same, this, by math instinct, is difficult to be true.
So we raise a nonconvergent example as follows. Define \[
x_n:= \int_0^{\sqrt{n}\pi } \sin x\,dx,
\] this is easily seen that $\{x_n\}$ satisifies the two conditions above, but $\{x_n\}$ diverges.$\qed$
Solution.
After a moment of thought one easily deduces that a sequence converges if and only if $\limn (x_{n+k}-x_n)=0$ uniformly in $k$. Therefore if every such sequence converges, the ``pointwise'' convergence in $k$ and ``uniform pointwise" convergence in $k$ will be the same, this, by math instinct, is difficult to be true.
So we raise a nonconvergent example as follows. Define \[
x_n:= \int_0^{\sqrt{n}\pi } \sin x\,dx,
\] this is easily seen that $\{x_n\}$ satisifies the two conditions above, but $\{x_n\}$ diverges.$\qed$
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