Getting probability greater than 1? Consider a random walk starting at a positive integer $k$. Now, trying to calculate the probability of returning to zero (from $k$), I did the following:
\begin{align}
P&(\text{returning to zero})  \\
&=P(\text{returning to zero in } k\text{ steps})  \\
&\quad+ P(\text{returning to zero in } k+2\text{ steps} )  \\
&\quad+ P(\text{returning to zero in } k+4\text{ steps} ) \\
&\quad+\cdots  \\
&=\Bigl({1\over 2}\Bigr)^k  \\
&\quad+\Bigl({1\over 2}\Bigr)^{k+2}{k+2\choose 2}  \\
&\quad+\Bigl({1\over 2}\Bigr)^{k+4}{k+4\choose 4}  \\
&\quad+\cdots  \\
&=\Bigl({1\over 2}\Bigr)^k\sum_{j=0}^\infty\Bigl({1\over 2}\Bigr)^{2j}{k+2j\choose 2j}
\end{align}
Now as I put some numbers in Desmos, it is very clear that this evaluates to greater than $1$. Clearly, I’ve made some mistake.

Clue :
$P$(Returning to zero in $k+2$ steps) should be $(\frac12)^{k+2}(k+1)$
Returning to zero in $k$ steps can only be done by taking the path $$k, k-1, \dots, 0$$
Returning to zero in $k+2$ steps can only be done by deviating exactly once from the '$k$ steps path'. So it's something like $$k, k-1, \dots, n, n+1, n, \dots, 0$$
There are $k+1$ integers between $0$ and $k$, leading to the result stated on top.
Now, if the walk is also allowed on the negatives, then there are $k+2$ possible deviations, because you can also go : $$k, \dots, 0, -1,0$$
But it's never ${k+2 \choose 2}$
I hope this was helpful.
