Does there exist a real differentiable function f with the following propertie simultaneously? Does there exist a real differentiable function f with the following property simultaneously?
if ‎put ‎$‎f_n:=f(n)‎$‎ and ‎‎$‎f^\prime_n:=f^\prime(n)‎$,‎$‎‎\mathbb{N}‎\subseteq ‎D_f‎$‎‎‎‎,then the sequence ‎$‎f_n‎$ ‎is convergent, ‎$‎f^\prime_n‎$‎ and ‎$‎\sum_{n=1}^{‎\infty‎}f^\prime_n‎$‎ are divergent but ‎‎‎‎‎‎ ‎$‎f^\prime_n+\sum_{k=1}^{‎n}f^\prime_k$‎‎ ‎is ‎convergent‎ as ‎$‎n\to {‎\infty‎}‎$.
 A: NO. Apply the following, with $f'(n)=B(n).$ 
Theorem. For a real sequence $(B(n))_n$ let $S(n)=\sum_{j=1}^n B(n)$ and $T(n)=B(n)+S(n).$  If $S(n)$ does not converge then $T(n)$ does not converge. Equivalently if $T(n)$ converges then $S(n)$ converges.
Proof: Suppose $L=\lim_{n\to \infty}T(n).$ By contradiction suppose that $(S(n))_n$ does not converge. Then (i) there exists $U>L$ such that $S(n)>U$ for infinitely many $n$, or (ii) there exists $U<L$  such that $S(n)<U$ for infinitely many $n.$ Case (ii) can be converted to Case (i) by replacing $B(n)$ with $-B(n)$ so it suffices to consider Case (i). 
Take $d>0$ with $3d<(U-L)$ and take $n_1$ such that $n\geq n_1\implies |T(n)-L|<d.$ Take $n_2\geq n_1$ such that $S(n_2)>U.$ We need the following Lemma: 
Lemma. There exists $n>n_2$ such that $S(n)<U.$ 
Proof of Lemma: By contradiction, suppose $S(n)>U$ for all $n\geq n_2.$ Then for every $j\geq 0$ we have $$L+d>T(n_2+j+1)=S(n_2+j)+2B(n_2+j+1)>U+2B(n_2+j+1)$$ implying  $B(n_2+j+1)<-(U-L-d)/2<-d.$ 
But then for positive integer $k>(S(n_2)-U)/d$ we have $S(n_2+k)=S(n_2)+\sum_{j=0}^{k-1}B(n_2+j+1)<S(n_2)-kd<U,$ contrary to $S(n_2+k)>U.$ 
So the lemma is proved.
By the Lemma let $n_3>n_2$ such that $S(n_3)<U.$ Now $S(n)>U$ for infinitely many $n$ so let $n_4$ be the $least$ $n>n_3$ such that $S(n)>U.$ Then $S(n_4-1)<U<S(n_4)=S(n_4-1)+B(n_4)$ so $$B(n_4)>0.$$ And we have $L+d>T(n_4)$ because $n_4>n_1.$  Therefore $$L+d>T(n_4)=S(n_4)+B(n_4)>S(n_4)>U>L+d$$ a contradiction.
