Can we partition the plane like this? 
Can a plane be written as union $\bigcup_{n=1}^{+ \infty}I_n$ of sets $I_n$ such that every $I_n$ is convex and an area of $I_n$ is $\frac{1}{n}$ and $I_n \bigcap I_m$ is a curve or an empty set if $m \neq n$?

Few minutes ago I remebered of harmonic series and started to think, because sum of it is $+ \infty$ can we in some way write a plane as union of infinite number of sets $I_n$, such that area of $I_n$ is $\frac {1}{n}$. There are such partitions of the plane when sets need not be convex, but what if we require that all are convex, like a question is, can we then do that?
Edit(bonus question):
What if we require that all convex sets have differentiable curves as boundaries?
 A: A sequence of rectangles $R_i$ of size $1\times a_i$ for $i=1,2,3\ldots$ with $a_i\ge 1$ can be filled with rectangles $1\times \frac{1}{n}$ generated by the harmonic series. Then the rectangles $R_i$ can be used to tile the plane by stacking them to form a sequence of parallel strips, each of width $1$. To avoid two tiles intersecting in a single point, shift every other strip by $\sqrt2$ in the direction of the strips.
A: It is hard to smoothly fit together closed convex sets with differentiable boundaries. It seems it might be possible to use only disks which sometimes intersect in a point (if that is allowed) or even are all disjoint. For the easier version let the $n$th disk have center as close to the origin as possible breaking ties by going counterclockwise. 
On second thought, this might be similar to an Apollonian Gasket and have (or be contained in a set of) fractal dimension.
A: Divide the plane into strips of width $1$. Partition each strip into rectangles with width $1$ and lengths (not in order) as follows:

Strip 1: $\frac11,\frac13,\frac15,\frac17,\cdots$
Strip 2: $\frac12,\frac16,\frac1{10},\frac1{14},\cdots$
Strip 3: $\frac14,\frac1{12},\frac1{20},\frac1{28},\cdots$
$\vdots$

Do you see the pattern? It is easy to prove that each strip can be covered by the specified rectangles, since $\frac1{2^{-k}}(\frac11+\frac13+\frac15+\cdots) = \infty$ and you can place the rectangles alternatingly on the left and right. It is also very basic number theory to prove that each reciprocal of a natural number occurs exactly once.
A: Here is a tiling using the rectangles $R_n$ of height $1$ and width $\frac1n$ with the feature that the location of any particular one is fairly explicit: 
The rectangles will form rows divided by the lines $y=k.$ 
Let $n=2^im$ with $m$ odd. Then the lower left corner of $R_n$ will be at $$(\frac{x_m}{2^i},k)$$ where $k=\frac{i}2$ or $k=-\frac{i+1}2$ according as $i$ is even or odd.
Above $y=0$ are the $R_m$ with odd $m$ in the order $$\cdots R_{11}R_7R_3R_1R_5R_9\cdots$$ with $R_1$ having lower left corner at $(0,0).$ This main row now determines the $x_m$ and the tiling. Specifically $x_m$ is either $1+1/5+1/9+\cdots+1/(m-4)$ or $-(1/3+1/7+1/11+\cdots+1/m).$
So to the right of the $y$-axis going up are $$\cdots R_{32}R_8R_2R_1R_4R_{16}\cdots$$ with each row the same as the main row except compressed horizontally by the appropriate power of $2.$ 
If desired, every other row can be shifted horizontally by $\sqrt{2}$ to avoid the points $(0,k)$ from being the intersections of two rectangles.
A: Concerning your "bonus question" about convex sets with differentiable boundaries, you may be interested in the old result by Helmut Groemer, On packings and coverings of unit density, Mathematika, 35 (1988), 172-179:

Let  $C_i$ be a sequence of convex bodies in  $\mathbb{R}^n$ of infinite total volume and whose diameters converge to 0. Then there exists a subsequence of $C_i$ that permits a packing of the unit cube $Q$ such that the set of points in $Q$ not covered by the packing is of measure zero.

It follows immediately that ${\mathbb R}^n$ can be almost partitioned into a sequence of smooth-boundary convex bodies whose total volume is infinite and whose diameters converge to zero. It seems that a perfect partition is impossible, so this may be the best we can expect.
A: Yes it can, even for the rectangles with vertical and horizontal sides. On $k$-th step the union of $I_1,\dots,I_{n_k}$ is a rectangle $[a_k,b_k]\times [c_k,d_k]$ where $a_k<-k<k<b_k,c_k<-k<k<d_k$. For doing next step, we add several rectangles to the top, then to the left, then to the bottom, then to the right so that new bounds exceed $k+1$ in absolute value. This is possible since the harmonic series diverges.
