What is the size of the marked area (calculate without computer)? Recently I scribbled a representation of squares which should challenge the imagination of the infiniteness of the sum $$s_1=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$$ and of the finiteness of the sum $$s_2=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...$$
The following picture puts together      


*

*the infinite length of a line constructed of the partial series of $s_1$ (using the length unit "cm" for example) and     

*the finite area constructed of the partial series of $s_2$ (in "cm²") which is enclosed by a border of infinite length:           


 
Ok, I mean this picture gives a nice composition (of a finite area limited by an infinite line) for some interesting amateur.            
After that, I thought: what if I smooth that figure by connecting the edges of the squares with straight lines? What is the size of the newly introduced area (yellow)? Is a purely geometrical solution possible? 

Challenge1: find a formula for the yellow area without the use of a computer. I think I needed about an hour to find the (very nice) solution with pen&paper (feeling a bit rusty needing such a long time... ).      
Challenge2: If the edges are connected with a smooth interpolating curve instead of straight pieces - what could be a formula with the target of simpleness and of straightness of the curve? (If I got the terminology correct, the ideal were a "absolute monotonous" function, see wikipedia.)
As a "poor man's approach" I thought one might rotate the picture by 45 deg first towards the x-axis. Then the new area could be expressed by the integral for the curve's formula. Give such an integral and the area.       
My first approach was to "generalize" the formula for polynomial interpolation using the heights of the upper edges (when the diagonal of the figure is rotated to the x-axis) towards an infinite number of interpolation-points; one problem with this is that the x-coordinates of the interpolation-points are not equidistant. (I've no solution so far, possibly one can employ the  $\psi()$-function as interpolation of the harmonic numbers.)
 A: I think I had also another, even better, solution for challenge1 than the following which I show below.
Here is the version using the telescoping sum.

Let's denote the yellow triangular pieces as $t_k$ beginning from the left, primarily looking above the main diagonal. We see, that each $t_k$ comes in two exemplars (above and below the diagonal) which can be combined to a rectangle of the size $A_k$. The area of the two $t_k$ is 
$$ A_k = \frac1{k(k+1)}$$
Then the partial series for the sequences of the yellow pieces (taken as pairs of same size) is
 $$\sum_{k=1}^n A_k = \frac12 + \frac16+\frac1{12}+ \cdots + \frac1{n(n+1)}$$
If we look at each term separately it can be decomposed in a difference of reciprocals:
 $$S(n)=\sum_{k=1}^n A_k = (1-\frac12) + (\frac12-\frac13)+(\frac13-\frac14)+ \cdots + (\frac1{n-1}-\frac1n)+(\frac1n-\frac1{n+1})$$
which gives by cancelletaion of the intermediate terms:
$$ S(n) = 1 - \frac1{n+1}$$ 
giving in the limit the area
$$ \lim_{n \to \infty} S(n) \text{ cm}^2  = 1 \text{ cm}^2 $$ 

The process from $S(n) $ to $S(n+1)$ can be pictured as follows: 


Here the yellow pieces are sequentially the pieces $t_k$ producing the partial sums
$$S(n)= A_1+A_2+...+A_n = 1 - \frac1{n+1} $$
by considering subsequently the residual areas. The area of the subsequent grey areas obviously go to zero.           
Remark: I think there are better graphic representations, but to start I give the first  one coming to mind.
A: If I am right, a smooth curve through the square corners are obtained via the digamma function in parametric form, as 
$$(\gamma+\psi_0(t),\gamma+\psi_0(t+1))$$ and
$$(\gamma+\psi_0(t+1),\gamma+\psi_0(t)).$$
A: This is an attempt to challenge2.              
This is the plot of the problem, where the upper and lower edges are connected by piecewise linear lines:

The better interpolation idea is, that the (somehow) "natural" interpolation of the harmonic numbers is the "digamma()" or "psi()"-function.
In Pari/GP we can implement the function "harmonic number" $h(x)$ as
$$ h(x) = \text{psi}(1+x)+\text{Euler} $$ to get for instance
$$ h(2) = 1.5 = 1+1/2 $$
I don't know any function being the functional inverse of $h(x)$ so I implement it using the solve() procedure to fake it at least for a certain domain.
$$ h^{o-1}(z) = \text{solve}(x=10^{-20},10^{50},h(x)-z) $$
Then I define
$$ f^+(x)= h(h^{o-1}(x)+1) \qquad \text{for the upper edges} \\ 
f^-(x)= h(h^{o-1}(x)-1) \qquad \text{for the lower edges} $$
With this we can plot the smooth interpolation $f(x)=f^+(x)$ based on the digamma()/psi()-function:

(I show here only a detail because the plot overlays the straight lines and the interpolation by $f(x)$ to partial invisibility)          
With this Pari/GP can even approximate the area between the upper interpolation and the main diagonal by the numerical computation of the integral as
$$ B = \int_{10^{-20}}^{50} f(t)-t \quad dt \approx  1.30075505601 $$
and the size of the yellow area (when its borders are interpolated by $f(x)$) is about
$$ 2 B - \zeta(2) \approx  0.956576045162 = 1 - 0.0434239548377 $$
which is about $0.0434239548377$ less than the value $1$ for the piecewise linear bordered yellow area.            
Whether this is the best, or at least most meaningful, solution I don't know. Especially I'd like to have a more common expression for $f(x)$ or at least a powerseries....          
