Coordinate bug

A good way to think of this problem is to find the $x$ and $y$ co-ordinates separately.

Look at the diagram,

Along the $x$ axis:

If we take a move rightward as positive and the move leftward as negative , then the bug takes the following moves:

$1, \frac{-1}{4}, \frac{1}{16}, \frac{-1}{64} ... \infty$

Notice that it forms an infinite geometric series with common ratio(r) = $\frac{-1}{4}$

Where the bug lands can be thought of as the sum of this infinite geometric series:

$Sum = \frac{a}{1-r}$

$\Rightarrow Sum = \frac{1}{1-\frac{-1}{4}} = \boxed{\frac{4}{5}}$

This gives us the $x$ coordinate.

Along the $y$ axis:

If we take a move upward as positive and the move downward as negative , then the bug takes the following moves:

$\frac{1}{2}, \frac{-1}{8}, \frac{1}{32}, \frac{-1}{128} ... \infty$

Notice again that it forms an infinite geometric series with common ratio(r) = $\frac{-1}{4}$

Again, where the bug lands can be thought of as the sum of this infinite geometric series:

$Sum = \frac{a}{1-r}$

$\Rightarrow Sum = \frac{\frac{1}{2}}{1-\frac{-1}{4}} = \boxed{\frac{2}{5}}$

This gives us the $y$ coordinate.

So, the point where it approaches is $\boxed{(\frac{4}{5}, \frac{2}{5})}$

Say the bug lives in the complex plane, and its path is $z_0,z_1,z_2,\ldots$ . We have $z_0=0$ and $z_{n+1}=z_n+\frac{1}{2^n} e^{in\frac{\pi}{2}}$

which looks ghastly, but encodes both the bug's turning (in the $e^{i n \frac{\pi}{2}}$ term) and the path shortening (in the $\frac{1}{2^n}$ term). This means we can find the bug's ultimate destination - $z_{\infty}$ - by summing a geometric series: $\begin{aligned} z_{\infty}&=\sum_{n=0}^{\infty} \frac{1}{2^n} e^{in\frac{\pi}{2}}\\ &= \frac{1}{1-\frac{1}{2} e^{i\frac{\pi}{2}}}\\ &=\frac{4}{5}+\frac{2}{5} i \end{aligned}$

Convert the $xy$ -plane into the complex plane of $z$ , where the real part $\Re(z) = x$ and imaginary part $\Im(z) = y$ . Let the first move be $u_0 = 1$ , the second move be $u_1 = \frac i2$ , third move be $u_2 = - \frac 14$ and so on. Then the final position of the bug after $n$ th move is given by $z_n = u_0+u_1+u_2 + \cdots + u_{n-1}$ . We note that a turning of $90^\circ$ anticlockwise in a complex plane is equivalent to multiplied by $i$ , the imaginary unit . Then we have:

$\begin{aligned} z_n & = \sum_{k=0}^{n-1} u_k = \sum_{k=0}^{n-1} \left(\frac i2 \right)^k \\ \implies \lim_{n \to \infty} z_n & = \sum_{k=0}^\infty \left(\frac i2 \right)^k = \frac 1{1-\frac i2} = \frac 2{2-i} = \frac {2(2+i)}5 = \frac 45 + \frac 25i \end{aligned}$

Therefore, the final coordinate of the bug is $\boxed{\left(\frac 45, \frac 25\right)}$ .

Since there are already multiple solution utilizing complex numbers, I'll give a solution using linear algebra.

Let's denote the first move by the vector $\mathbf{b} = \begin{bmatrix} 1 \\ 0\\ \end{bmatrix}.$

Now counterclockwise rotation of $90$ degrees corresponds to multiplying by the matrix $\mathbf{M} = \begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix},$

and halving the length corresponds to multiplying by the matrix

$\mathbf{N} = \begin{bmatrix} \frac12 & 0\\ 0 & \frac12 \\ \end{bmatrix}.$

Let's denote their product by $\mathbf{NM} = \mathbf{A} = \begin{bmatrix} 0 & -\frac12\\ \frac12 & 0\\ \end{bmatrix}.$

Now we can get the next vector to add by multiplying the previously added vector by matrix $\mathbf{A}$ .

Continuing this indefinitely we see that the sum $\mathbf{x} = \sum\limits_{k=0}^{\infty} \mathbf{A}^k \mathbf{b}$ gives the limit point if it converges.

Since the eigenvalues of the matrix $\mathbf{A}$ , which are $\pm \dfrac i2$ , have absolute values less than $1$ , the geometric series $\sum\limits_{k=0}^{\infty} \mathbf{A}^k$ converges towards $\left(\mathbf{I} - \mathbf{A}\right)^{-1}$ and thus

$\mathbf{x} = \left(\mathbf{I} - \mathbf{A}\right)^{-1} \mathbf{b} = \left( \begin{bmatrix} 1 & 0\\ 0 & 1 \\ \end{bmatrix} - \begin{bmatrix} 0 & -\frac12\\ \frac12 & 0\\ \end{bmatrix} \right)^{-1} \begin{bmatrix} 1 \\ 0\\ \end{bmatrix} = \begin{bmatrix} 1 & \frac12\\ -\frac12 & 1 \\ \end{bmatrix}^{-1} \begin{bmatrix} 1 \\ 0\\ \end{bmatrix} = \begin{bmatrix} \frac45 & -\frac25\\[6pt] \frac25 & \frac45 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0\\ \end{bmatrix} = \begin{bmatrix} \frac45 \\[6pt] \frac25 \\ \end{bmatrix}.$

Thus the limit point is $\boxed{\left(\dfrac45, \dfrac25\right)}$ .