MHB What point does the spiral converge to?

[amr21]

Starting from the origin, go one unit east, then the same distance north, then (1/2) of the previous distance west, then (1/3)
of the previous distance south, then (1/4) of the previous distance east, and so on. What point does this 'spiral' converge to?

I have attempted to sketch this out but not sure how to work out what point it converges to. I know that it is somewhere between n/2 and 7n/12 (after 6 moves from the origin). The fractions are getting smaller but is there an easy way to work out at what point it converges?

 

[I like Serena]

Starting from the origin, go one unit east, then the same distance north, then (1/2) of the previous distance west, then (1/3)
of the previous distance south, then (1/4) of the previous distance east, and so on. What point does this 'spiral' converge to?

I have attempted to sketch this out but not sure how to work out what point it converges to. I know that it is somewhere between n/2 and 7n/12 (after 6 moves from the origin). The fractions are getting smaller but is there an easy way to work out at what point it converges?

Hi amr21,

Looking at the X coordinate, we have:
$$X=1-\frac 12 + \left(\frac 12 \cdot\frac 13 \cdot\frac 14\right) - \left(\frac 12 \cdot\frac 13 \cdot\frac 14 \cdot\frac 15 \cdot\frac 16\right) + ...$$
This looks a bit like the power series of the cosine:
$$\cos x = 1 - \frac 1{2!} x^2 + \frac 1{4!} x^4 - ...$$
Can we find an $x$ to match them? (Wondering)

 

[amr21]

Hi amr21,

Looking at the X coordinate, we have:
$$X=1-\frac 12 + \left(\frac 12 \cdot\frac 13 \cdot\frac 14\right) - \left(\frac 12 \cdot\frac 13 \cdot\frac 14 \cdot\frac 15 \cdot\frac 16\right) + ...$$
This looks a bit like the power series of the cosine:
$$\cos x = 1 - \frac 1{2!} x^2 + \frac 1{4!} x^4 - ...$$
Can we find an $x$ to match them? (Wondering)

That makes sense, thank you! Could you explain further on how to find an x to match them?

 

[I like Serena]

That makes sense, thank you! Could you explain further on how to find an x to match them?

We should match $-\frac 12$ with $-\frac 1{2!}x^2$, implying that $x^2=1$, which means that $x=1$ will do the trick.
It means that the spiral ends at $X=\cos(1)$.