D: HALP GAIZ!

(Sorry if I'm not supposed to post homework here or something.)

My Analysis prof gives hard problems as extra-credit sometimes, and he gave this one during our first class of the semester yesterday.  I can't get anywhere with it. :(  It looks like it should be totally easy, but I'm not seeing it.  I just have to prove that the sequence {sin(1), sin(4), sin(9), ...} doesn't converge. The numbers 1,4,9... are the integer squares.

Anyone have any ideas?

Write

x \equiv y \pm \delta (\text{mod}\;\pi)

to mean that there is some number in [x-(y+\delta),x-(y-\delta)] which is an integral multiple of \pi.  Let L be a hypothetical limit of sin(n^2) and let c,c' be the two solutions in [0,2\pi) of sin(x)=L.  Note that c' \equiv -c (\text{mod}\;\pi).

With this notation, if the limit exists, then for any \delta > 0, there is an N such that for all n \ge N we have

n^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+1)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+2)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+3)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+4)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
(n+5)^2 \equiv \pm c \pm \delta (\text{mod}\;\pi)
...

(Note this actually includes the solutions of sin(x) = -L, but that doesn't matter).  Subtract the first of these from the second, the second from the third, and so on to get

2n+1 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+3 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+5 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+7 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
2n+9 \equiv \pm c \pm c \pm 2\delta (\text{mod}\;\pi)
...

for some choice of the signs.  Now subtract the first of these from the rest to get:

0 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
2 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
4 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
6 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
8 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
10 \equiv \pm c \pm c \pm c \pm c \pm 4\delta (\text{mod}\;\pi)
...

Ignoring the error term, there are only 5 possibilities for the RHS of this last set of equations: \{-4c, -2c, 0, 2c, 4c\}.  Therefore two of these equations have the same RHS.  Of these two, subtract the one with the smaller LHS from the other to get an equation like

a \equiv \pm 8\delta (\text{mod}\;\pi)

for an even integer 2 \le a \le 10.  Contradiction for small enough \delta.