# What point does the spiral which starts at the point $0$, go up $1$, then right $1/2$, then down $1/3$, then left $1/4$, then up $1/5$ converge to? [duplicate]

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?

• Deal with the $x$ component separately from the $y$ component. Write down series for each. – David G. Stork Jun 1 at 1:12
• @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 $\mathbb{R}$. – Xander Henderson Jun 18 at 1:15
• @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. – Xander Henderson Jun 18 at 1:18
• @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. – Integrand Jun 18 at 1:18
• @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). – Xander Henderson Jun 18 at 2:35

# 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

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

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