By elementary means:
In the figure below, we have drawn a quarter circle inside a unit square and divided the top side in $n$ equal parts. Then we have chosen the height of the blue rectangles in such a way that the area of a rectangle equals the area of the corresponding sector (the aperture angles go decreasing).

If you look at the enhanced triangle, its long sides have length $\sqrt{1+x^2}$, by Pythagoras, so that its short side is $\sqrt{1+x^2}$ times longer than the arc on the circle ($x$ horizontal in $[0,1]$). Then by similarity of triangles, the ratio of the width of a rectangle to this short side is also $\sqrt{1+x^2}$, and we conclude that the equation of the green curve is
$$y(x)=\frac1{2(1+x^2)}.$$
Furthermore, the area under the curve is the same as the area of the eighth of a circle, i.e. $\dfrac\pi8$.
Now the (doubled) area is obtained as the sum of the areas of the rectangles,
$$\frac\pi4=2\sum_{k=1}^n\frac1ny\left(\frac kn\right)=\frac1n\sum_{k=1}^n\frac1{1+\dfrac{k^2}{n^2}}.$$
To evaluate it $^{(1)}$, we use the identity
$$\frac1{1+\dfrac{k^2}{n^2}}=1-\frac{k^2}{n^2}+\frac{k^4}{n^4}-\frac{k^6}{n^5}+\cdots$$
and by the Faulhaber formula $^{(2)}$, we know that the sum of the $n$ first $p^{th}$ powers is $\dfrac{n^{p+1}}{p+1}$.
This gives
$$\frac\pi4=\frac1n\left(n-\frac{n^3}{3n^2}+\frac{n^5}{5n^4}-\frac{n^7}{7n^6}+\cdots\right)$$ which is the claimed formula.
$(1)$ This is "justified" by
$$(1+x)(1-x+x^2-x^3+x^4-\cdots)=1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5\cdots=1\pm x^\infty$$ where the last term vanishes when $x<1$.
$(2)$ This is "justified", using the binomial theorem, by
$$n^p=\sum_{k=1}^n k^p-\sum_{k=1}^{n-1} k^p\approx\frac{n^{p+1}}{p+1}-\frac{(n-1)^{p+1}}{p+1}=\frac{n^{p+1}-(n^{p+1}-(p+1)n^p+\binom{p+1}2n^{p-1}-\binom{p+1}3n^{p-2}+\cdots)}{p+1}=n^p+\cdots.$$
The neglected terms are of lower degree in $n$ and can be ignored for large $n$.