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In the plane, start at the point $0$, go up $1$, then right $1/2$, then down $1/3$, then left $1/4$, then up $1/5$, etc. This is a spiral with infinite length because the harmonic series diverges. But when I draw the picture it looks like the spiral converges to a point. What are the coordinates of that point?

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  • $\begingroup$ Deal with the $x$ component separately from the $y$ component. Write down series for each. $\endgroup$ – David G. Stork Jun 1 at 1:12
  • $\begingroup$ @Integrand I don't think that the linked question answers this question: this question is about the limit of a sequence of points in $\mathbb{R}^2$, while the related question is about the convergence of a sequence in $\mathbb{R}$. $\endgroup$ – Xander Henderson Jun 18 at 1:15
  • $\begingroup$ @Francis Ray You might think about treating the $x$- and $y$-coordinates separately. For example, the sequence of $x$-coordinates is $0$, $0$, $1/2$, $1/2$, $1/4$, $1/4$, etc. It should be pretty straight-forward to determine the general term and show that it converges. $\endgroup$ – Xander Henderson Jun 18 at 1:18
  • $\begingroup$ @Xander Henderson, 'spiral' suggests $\mathbb{R}^2$ and aside from being off by one index, it's exactly he same question. Helpful for OP to compare their question with the one I linked. $\endgroup$ – Integrand Jun 18 at 1:18
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    $\begingroup$ @Integrand That being said, I do agree that the questions are very closely related, and that a fairly mature mathematician should quickly see the link. I just think that there is enough daylight between the two to warrant keeping this question open (or, at least, to close it for lacking context rather than as a duplicate). $\endgroup$ – Xander Henderson Jun 18 at 2:35
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Setup

So, consider the series of points we go to along the way:

  • $(x,y) = (0,1)$
  • $(x,y) = (1/2,1)$
  • $(x,y) = (1/2,1-1/3)$
  • $(x,y) = (1/2-1/4,1-1/3)$
  • $(x,y) = (1/2-1/4,1-1/3+1/5)$

The pattern seems obvious at this point. The point where you'll end up is given by

$$(x,y) = \left( \frac 1 2 - \frac 1 4 + \frac 1 6 - \cdots , 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \cdots \right)$$

In sigma notation,

$$(x,y) = \left( \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} , \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \right)$$


Finding the $x$-coordinate

Now, recall, the series for the natural logarithm (or, rather, one of several series):

$$\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n x^k$$

We note that, with evaluation at $x=1$,

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} = \frac 1 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \frac{\ln(2)}{2}$$


Finding the $y$-coordinate

Next, recall the series for arctangent:

$$\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1}$$

Evaluation at $x=1$ gives us this:

$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \arctan(1) = \frac{\pi}{4}$$


Conclusion

Thus, the point you are arrive at is given by

$$(x,y) = \left( \frac{\ln(2)}{2} , \frac{\pi}{4} \right) \approx (0.347,0.785)$$

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The $x$ coordinate is $$\sum\limits_{n=1}^\infty(-1)^{n+1}\frac1{2n}=\frac{\ln(2)}2$$The $y$ coordinate is$$\sum\limits_{n=0}^\infty(-1)^n\frac1{2n+1}=\frac\pi4$$

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