Is $f: F \to R, \ (a_j)_{j \in \mathbb N} \mapsto \sum_{j \in \mathbb N} \ a_j $ bijective and find the inverse function! $F$ is the set of the sequences in $\mathbb C$ and $R$ is the set of the series in $\mathbb C$.
$f: F \to R, \ (a_j)_{j \in \mathbb N} \mapsto \sum_{j \in \mathbb N} \ a_j $
Now $\sum_{j \in \mathbb N} \ a_j $ should be the sequence of the partial sums $(s_n)_{n \in \mathbb N}$ with $s_n =\sum_{j \in \mathbb N} \ a_j $


*

*Show this transformation is BIJECTIVE!

*Find the inverse function!

Thoughts:


*

*Because $F$ is the set of the sequences and set of the series $R$ is the sequence of the partial sums => $F=R$

*So $f: F \to F, f(a)=y$ with $y_n:=\sum_{j=1}^n  \ a_j$ for $a= (a_j)_{j \in \mathbb N}$

*inverse function: $f^{-1}: F \to F,f^{-1}\sum_{j=1}^n  \ a_j\mapsto R(a_i)$ with $a_i\sum_{j=1}^i a_i -\sum_{j=1}^{i-1} a_i$



Questions:


*

*How can I proove that it is bijective? (I think the sets F,R have the same cardinality)

*Is the inverse function correct?
 A: As @Hagen von Eitzen commented, your inverse function is correct, and you can conclude that it is bijective because there is an inverse. Yet I would like to point out a pair of similar operators which might be quite known to you.
Let $D : C^\infty \to C^\infty$ be an operator on the set of infinitely-differentiable functions given by:
$$D(f) = f' = x\mapsto \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
and set $I : C^\infty \to C^\infty$ be defined as
$$I(f) = x\mapsto\int_{[0,x)} f(t)\ \mathrm{d}t$$
Then, $D\big(I(f)\big) = f$ and $I\big(D(f)\big) = x\mapsto f(x) - f(0)$ (thanks to @Berci for correction; also, for convenience, we could consider only functions with $f(0) = 0$). Does these two seem familiar?
Now, if we were to adapt these operators to functions $f : \mathbb{N} \to \mathbb{R}$, we could write:
\begin{align}
\Delta(f) &= n\mapsto\frac{f(n+1)-f(n)}{1}\\
S(f) &= n \mapsto \sum_{k \in [0,n)\cap\mathbb{N}} f(k) = n \mapsto \sum_{k = 0}^{n-1} f(k)
\end{align}
assuming convention where $\sum_{k=0}^{-1} f(k) = 0$. Simple calculation shows that $\Delta\big(S(f)\big) = f = S\big(\Delta(f)\big)$. For more, see here.
Now there is a very short step from $\Delta$ and $S$ to the two functions of yours.
I hope this helps $\ddot\smile$
