Does there exist a real differentiable function $f$ with the following properties? (a) ‎$‎‎\mathbb{N}‎\subseteq D_f‎‎$‎
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(b) If we put ‎$‎f_n := f(n)‎$‎ and ‎$‎f'_n := f^\prime(n)$, then the sequences ‎$‎f_n‎$ ‎and ‎‎$f'_n+\sum_{k=1}^n f'_k‎$ ‎are ‎convergent ‎but ‎‎$‎f'_n‎$ ‎is ‎divergent ‎as ‎‎$‎n‎\rightarrow ‎\infty‎$‎.‎
 A: Let $(a_n:n\in\Bbb N)$ be a real sequence such that the limit $a:=\lim_{n\to\infty}\left(2a_n+\sum_{k=0}^{n-1}a_k\right)$ exists. Then, given $\epsilon>0$, there exists $m\in \Bbb N$ such that $$a-\epsilon<2a_n+\sum_{k=0}^{n-1}a_k<a+\epsilon$$whenever $n>m$. Similarly, $$a-\epsilon<2a_{n+1}+\sum_{k=0}^na_k<a+\epsilon.$$By subtraction and halving,$$-\epsilon<a_{n+1}-\tfrac12a_n<\epsilon.$$Likewise (but not halving),$$-2\epsilon<2a_{n+2}-a_{n+1}<2\epsilon.$$Continuing, and scaling up by $2$ each time, we get$$-4\epsilon<4a_{n+3}-2a_{n+2}<4\epsilon$$and so on until$$-2^{k-1}\epsilon<2^{k-1}a_{n+k}-2^{k-2}a_{n+k-1}<2^{k-1}\epsilon.$$Adding all these up gives$$-(2^k-1)\epsilon<2^{k-1}a_{n+k}-\tfrac12a_n<(2^{k-1}-1)\epsilon.$$Now divide through by $2^{k-1}$ to get$$-2\epsilon<a_{n+k}-2^{-k}a_n<2\epsilon.$$By taking $k>\log_2|a_n/\epsilon|$, we obtain $$-3\epsilon<a_{n+k}<3\epsilon.$$It follows that $$\lim_{n\to\infty}a_n=0.$$Hence the divergence of $f'(n)$ ($=a_n$) is inconsistent with the other conditions given in the question, and so no such function $f$ can exist.
