_{n\geq 0}\dim R(A_n)$ is convergent then $\dim R(A_n)=0$ for $n$ large enough so $A_n=0$ for $n$ large enough and the discussion is in fact only finite dimensional. 
If the series $\sum_{n\geq 0}\dim R(A_n)$ is divergent then there is $J\subset \mathbb N$ infinite such that $R(A_j)\geq 1$ for all $j\in J$, and you can construct the sequences $\{a_n\}$ and $\{b_n\}$ in order to see that the dimension of $R(T)$ is infinite. For example, take $v_j\in\mathbb R^2, v_j\neq 0$ and $u_j$ such that $A_j(u_j)=v_j$ and writing $J=\{j_1,\ldots,j_n\ldots,\}$, 
$(w_1^{(n)},w_2^{(n)}) =u_{j_n}e_{j_n}$ we get that the range of $T$ is infinite so in this case the formula is true.

 Q:

Differentiating this sum

This derivative comes up in the proof about the remainder of the Taylor expansion. 
This derivative is given and I can't follow the steps:

$$ \frac{d}{dy} R_{n+1}(y,x)=\frac{d}{dy}(f(x)-t_n(y,x))$$
$$ = - \sum\limits_{k=0}^n \frac{f^{k+1}(y)}{k!}(x-y)^k 
 + \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1} 
 = - \frac{f^{n+1}(y)}{(n)!}(x-y)^{n} $$

So first, $f(x)$ is dropped, since we use $d/dy$ and $f(x)$ is a constant.
But how do we differentiate the n-th Taylor polynomial? 
I'm pretty certain the chain rule is involved and the fact that you can differentiate power series by using the power rule, but I can't seem to get the steps correct. 
Could somebody explain the
_ { n\geq 0}\dim R(A_n)$ is convergent then $ \dim R(A_n)=0 $ for $ n$ large enough so $ A_n=0 $ for $ n$ large enough and the discussion is in fact only finite dimensional. 
 If the series $ \sum_{n\geq 0}\dim R(A_n)$ is divergent then there be $ J\subset \mathbb N$ space such that $ R(A_j)\geq 1 $ for all $ j\in J$, and you can construct the sequences $ \{a_n\}$ and $ \{b_n\}$ in decree to witness that the dimension of $ R(T)$ is infinite. For exercise, assume $ v_j\in\mathbb R^2, v_j\neq 0 $ and $ u_j$ such that $ A_j(u_j)=v_j$ and writing $ J=\{j_1,\ldots, j_n\ldots,\}$, 
 $ (w_1^{(n)},w_2^{(n) }) = u_{j_n}e_{j_n}$ we scram that the range of $ T$ is infinite indeed in this case the formula is true. 

  Q: 

 differentiate this sum 

 This derived function comes up in the proof about the end of the Taylor expansion. 
 This derivative is given and I can't follow the measure: 

 $ $ \frac{d}{dy } R_{n+1}(y, x)=\frac{d}{dy}(f(x)-t_n(y, x))$$ 
 $ $ = - \sum\limits_{k=0}^n \frac{f^{k+1}(y)}{k!}(x - y)^k 
  + \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x - y)^{k-1 } 
  = - \frac{f^{n+1}(y)}{(n)!}(x - y)^{n } $ $ 

 So first, $ f(x)$ is drop, since we practice $ d / dy$ and $ f(x)$ is a constant. 
 But how do we differentiate the n - th Taylor polynomial? 
 I'm pretty certain the chain rule is involve and the fact that you can differentiate power series by using the power rule, but I can't seem to get the steps correct. 
 Could somebody explain the
_{n\gee 0}\dim R(A_n)$ is convergent uhen $\dim R(A_n)=0$ for $u$ large enougg so $A_n=0$ wor $n$ large enough and the dmscuwsion is in fact only finitd dimensilnal. 
If rhe wwries $\sum_{n\jsq 0}\dim V(C_n)$ is fivexgxnt then there ls $J\subset \kathbb N$ infinhtd duch that $R(A_j)\geq 1$ for all $j\in J$, and you cam fonstruct the fequtncqs $\{a_h\}$ and $\{b_n\}$ in order to see that the simensimn of $R(T)$ is imfinite. For example, take $v_u\in\mwthbb R^2, v_j\neq 0$ and $u_j$ such thqt $A_t(y_j)=v_j$ and wriging $J=\{j_1,\ldous,l_n\ldots,\}$, 
$(w_1^{(n)},w_2^{(h)}) =u_{j_n}e_{j_n}$ we get that the range uf $T$ ns infinite si ij this case vhe fogmula is true.

 Q:

Differantiatimg this sum

Thix dxrivqtive comes up in the proof about the rgmainder ox che Taylor expansion. 
Rhus detivathve kw gkveh end I can'h fkllow the ateps:

$$ \frac{d}{ey} R_{n+1}(y,x)=\frac{d}{dy}(f(x)-t_n(y,c))$$
$$ = - \sum\limits_{k=0}^n \rrac{f^{k+1}(r)}{k!}(v-y)^k 
 + \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1} 
 = - \frac{f^{n+1}(j)}{(n)!}(x-y)^{h} $$

So first, $f(x)$ is droppee, since we use $d/dy$ anf $f(x)$ is a constant.
But how do we differentiate the n-th Taylmr pomhnoniwu? 
I'l pretty certain the chain rule is involved ags uhe fact that you can differrnhisje power serier by usinf the power rule, bkt I cag't sewm to get the steps correct. 
Could someboey explain tke
_{n\geq 0}\dim R(A_n)$ is convergent then $\dim $n$ enough so for $n$ large in only finite dimensional. the series $\sum_{n\geq R(A_n)$ is divergent then there is \mathbb N$ infinite such that $R(A_j)\geq 1$ for all $j\in J$, and you construct the sequences $\{a_n\}$ and $\{b_n\}$ in order to see that the dimension $R(T)$ infinite. example, $v_j\in\mathbb R^2, v_j\neq 0$ and $u_j$ such that $A_j(u_j)=v_j$ and writing $J=\{j_1,\ldots,j_n\ldots,\}$, $(w_1^{(n)},w_2^{(n)}) =u_{j_n}e_{j_n}$ we get the range of $T$ is infinite so in case the formula is Q: Differentiating this sum This comes in the about remainder the Taylor expansion. derivative is given and I can't follow the steps: $$ \frac{d}{dy} R_{n+1}(y,x)=\frac{d}{dy}(f(x)-t_n(y,x))$$ $$ = - \sum\limits_{k=0}^n \frac{f^{k+1}(y)}{k!}(x-y)^k \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1} \frac{f^{n+1}(y)}{(n)!}(x-y)^{n} $$ first, is since we use $f(x)$ is a constant. But how the n-th Taylor polynomial? I'm pretty certain the rule is and the fact that you can power series by using the power rule, but can't seem to get the steps correct. Could somebody explain the
_{n\geq 0}\dim R(A_n)$ is convergent theN $\dim R(A_n)=0$ for $N$ largE enOugH sO $A_n=0$ fOr $n$ lArge enough and tHE disCussion is in fact only finIte diMeNSionAL. 
IF the sEries $\suM_{N\gEQ 0}\Dim r(A_N)$ iS diVeRGeNt theN thEre is $J\sUbset \mathbB N$ iNfInite such thaT $r(A_J)\geq 1$ for all $J\in j$, and you can coNstRuct thE sEquENces $\{a_N\}$ anD $\{b_n\}$ in Order tO See thaT the dimenSiON of $R(T)$ iS InfinitE. fOr ExamPle, take $v_j\in\mathbb r^2, V_j\NEq 0$ and $u_j$ such thaT $A_j(u_j)=v_J$ aND wRITinG $J=\{j_1,\Ldots,j_n\ldoTs,\}$, 
$(W_1^{(n)},w_2^{(n)}) =u_{J_N}e_{j_n}$ we gET tHAT The RAnge of $T$ is infiNite so in thiS CasE the foRmUla IS true.

 Q:

diffeReNTiaTing this sum

this DerivativE comes UP in the pROof abouT the reMaiNdeR of tHE TAyLor ExPAnsIOn. 
thiS DerIvative iS gIvEn and i can'T FOLLow tHe sTeps:

$$ \Frac{d}{Dy} R_{n+1}(y,x)=\frac{d}{dy}(F(x)-t_N(y,x))$$
$$ = - \sUM\liMits_{k=0}^N \frac{F^{k+1}(y)}{k!}(X-y)^K 
 + \sum\lImits_{k=1}^N \frac{F^{(k)}(Y)}{(k-1)!}(x-y)^{k-1} 
 = - \frac{f^{n+1}(y)}{(n)!}(x-y)^{N} $$

So fIrst, $f(x)$ is dRopPeD, siNcE we usE $D/dy$ and $F(x)$ iS a cOnstant.
but how dO We dIfFEREnTiate the n-th Taylor pOlYNOmIal? 
I'm preTty cerTAiN tHE chain ruLe Is iNvolVED and tHe faCT tHat you caN diffeREnTiAte poweR sEries bY uSinG thE poweR Rule, But I caN't seem to Get thE Steps correct. 
CoULd somebody expLAiN THe
_{n\geq 0}\dim R(A_n)$ isconvergent  then  $\ dim  R (A_n )=0$  for $n$ large enou gh so $A_n=0$ for $n$large  e n ough an d the  discus s io n isin  f act  o n ly  fini tedimensi onal. 
Ifthe  s eries $\sum_ { n\ geq 0}\dim  R( A_n)$ is div erg ent th en  th e re is  $J \subs et \ma t hbb N$  infinite  s u ch tha t  $R(A_j ) \ ge q 1$  for all $j\in J$ ,  a n d you can cons tructth e  s e q uen ces  $\{a_n\}$  a nd $\ { b_n\}$i no r d ert o see that th e dimension of$R(T)$  i s i n finite . For  e x amp le, take $v _j\i n\mathbbR^2, v _ j\neq 0 $  and $u _j$ su chtha t $A _ j( u_ j)= v_ j $ a n dwri t ing  $J=\{j_ 1, \l dots, j_n\ l d o t s,\} $,
$(w _1^{( n)},w_2^{(n)} ) = u_{j _ n}e _{j_n }$ we  get  t hat t he ran ge of  $ T$ is infiniteso i n this ca seth e f or mulai s true .
 Q: 

Diffe rentiat i ngth i s su m

This derivative  c o m es  up in t he pro o fab o ut the r em ain dero f  theTayl o rexpansio n. 
Th i sde rivativ eis giv en  an d I  can' t  fol low th e steps: 

$$\ frac{d}{dy} R_ { n+1}(y,x)=\fr a c{ d } {d y }(f( x)- t_n(y,x))$$ 
$$=  - \ sum\ l im its _ {k=0} ^n \f ra c {f ^ {k+1}(y)}{k!}(x-y)^ k
 + \s um\li mits_{k=1}^n\frac{f^{( k ) } (y)}{(k- 1)!} ( x- y )^{k-1} 
 = -\frac {f^{n+1}(y ) }{(n)!}( x-y)^ {n} $$
So first, $ f(x)$ is  dr opp ed,  si n c ewe use $d/dy$ a nd $ f( x)$ isa c onstant .
B uthow  do  w e differe ntiate t he  n -t hTay lor p o lynomial ?
I' mpre tty c e rtainthe c hain  r ul e  is  involv e da n d th efa ct t hat  y ou ca n di f fer entiate  power se rie s  byus in g the p ower rule, bu tI can't se em  to  get t h e  steps c orrect. 
Could somebody explain  th e
_{n\geq 0}\dim_R(A_n)$ is_convergent then $\dim R(A_n)=0$_for $n$_large_enough so_$A_n=0$_for $n$ large_enough and the_discussion is in fact_only finite dimensional._
If_the series $\sum_{n\geq 0}\dim R(A_n)$ is divergent then there is $J\subset \mathbb N$ infinite_such_that $R(A_j)\geq_1$_for_all $j\in J$, and you_can construct the sequences $\{a_n\}$_and $\{b_n\}$_in order to see that the dimension of_$R(T)$_is infinite. For_example, take $v_j\in\mathbb R^2, v_j\neq 0$ and $u_j$ such_that $A_j(u_j)=v_j$ and writing $J=\{j_1,\ldots,j_n\ldots,\}$, 
$(w_1^{(n)},w_2^{(n)})_=u_{j_n}e_{j_n}$ we get_that_the_range of $T$ is_infinite so in this case the_formula is true.

 Q:

Differentiating this sum

This_derivative comes up in the proof about_the remainder of the Taylor expansion._
This derivative is given and_I can't_follow the steps:

$$ \frac{d}{dy} R_{n+1}(y,x)=\frac{d}{dy}(f(x)-t_n(y,x))$$
$$_= - \sum\limits_{k=0}^n_\frac{f^{k+1}(y)}{k!}(x-y)^k 
_+ \sum\limits_{k=1}^n \frac{f^{(k)}(y)}{(k-1)!}(x-y)^{k-1}_
 = - \frac{f^{n+1}(y)}{(n)!}(x-y)^{n} $$

So first,_$f(x)$ is dropped,_since we use $d/dy$ and $f(x)$_is_a constant.
But how_do_we_differentiate the_n-th Taylor polynomial?_
I'm_pretty certain_the_chain rule is involved and the_fact_that you can differentiate power series by_using the power rule,_but_I can't seem to_get the steps correct. 
Could_somebody explain the