Relationship between the graphs of $nCk$ and continuous functions I was curious about functions that involve $nCk$ namely "n choose k" so I just thought of this function:
$$ f(x) = \binom{x+2}{x}^{-1}$$
This function looks like this:

And I thought what is the integral of this on the interval $[0, \infty)$
$$ \int_0^{\infty} \binom{x+2}{x}^{-1} dx$$
It looks like a lot of rectangular areas together. However if we use the fact that:
$$ nCk = \binom{n}{k} =\frac{n!}{k!(n-k)!} \\ \binom{x+2}{x}^{-1} = \frac{x!(x+2-x)!}{(x+2)!} = \frac{2x!}{x!(x+1)(x+2)} = \frac{2}{(x+1)(x+2)}$$
Which is a continuous function that looks like:

And together on the same grid:

Question: Is there a relationship between these? It does not seem like the integral
$$ \int_0^{\infty} \frac{2}{(x+1)(x+2)} dx$$
Is the same as the prior integral, but the functions are the same aren't they?
I would like to hear your ideas! Thank you :-)

You are right, it is because, a priori, the binomial number is defined for integer valued numbers. What is probably happening in your CAS is that they are rounding to the nearest integer the input number (denoted by $\lfloor x\rceil =\arg \min _{n}\{|n-x|\},$ for example $\lfloor 7.3\rceil=7,\lfloor 7.6\rceil=8$). So the orange is actually:
$$\frac{2}{(\lfloor x \rceil+1)(\lfloor x \rceil+2)}.$$ which is not the same function.If you want the area for the orange, notice that is the sum of the areas of those rectangles of width $1$ and height $\frac{2}{(n+1)(n+2)}=2\left (\frac{1}{n+1}-\frac{1}{n+2}\right ).$ You can use the telescopic sum to find your integral.
If, insted, you want the area of the blue one, a similar decomposition shows the presence of a logarithm.

It is worth noticing that the area of the orange is an approximation of the area of the blue one, by construction of Riemann integral.
