I’m trying to check if a graph with $40$ vertices with $deg(v_1) = 1$, $deg(v_2)= 2$, $\cdots$ , $deg(v_{40}) = 40$ exists. No other characteristics are known about this graph.
I suppose that a graph like that can’t exist simply because the vertex $40$ cannot have degree $40$. Is my assumption correct or is strictly referred to the hypothesis of a connected graph and it’s not enough?
It depends on the kind of a graph. If it's a simple graph, your argument is absolutely correct. But, if it's not a simple graph, and there can be multiple edges between 2 vertices or loops, then we can have this graph. One of the many possible graphs in that case can be:-
(Note that i'll take $deg(v_i)=i$)
$v_1$ will be connected to $v_2$ with one edge. $v_{2i}$, where $i$ is integer, is connected to $v_{2i-1}$ and $v_{2i+1}$ with i edges each. $v_{2i+1}$, where $i$ is integer, is connected to $v_{2i}$ via $i$ edges and with $v_{2i+2}$ via i+1 edges. Finally, $v_{40}$ will be connected to $v_{39}$ via $20$ edges. It can then have $10$ self loops, with each loop contributing $2$ to its degree, giving $deg(v_{40})=40$
EDIT:
Since it was asked to do it do it without self-loops, we can do it in this way:-
Connect (1) $$v_{40}->v_{1}, v_2,\cdots ,v_{39} $$ $$v_{39}->v_{2}, v_3,\cdots ,v_{38} $$ $$v_{38}->v_{3}, v_4,\cdots ,v_{37} $$ $$\cdots $$ $$v_{22}->v_{19}, v_{20}, ,v_{21} $$ $$v_{21}->v_{20}$$
Then (2) Join $v_{20+2i}$ to $v_{20+2i-1}$.
In this way, for $v_i$, $i\leq 20$, it'll come only i times in the above mentioned "joining". For $v_{20+i}$, $1\leq i\leq 20$, it'll be connected to vertices $v_{20-i+1}$ to $v_{20+i-1}$ via pairing (1) $\implies$ contributes $2(i-1)+1$ to degree. Also, it also connected to all the vertices above it, again via (1) pairing $\implies$ contributes $40-20-i$ to degree. And finally, another vertex via (2) pairing $\implies$ contributes $1$ to degree. Final degree $=(2i-2+1)+(40-20-i)+1=20+i$.