n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$
If it's not, then yay, I found it. But I think it's very unlikely that it's not already known, as with anything else I've found out.
Edit:
All right, I just checked the answer with the Polygamma on Mathematica, and it doesn't give it a closed-form, but I do have and am adding it here.
Please don't vandalize the rating of this post otherwise I won't be able to add the details.

$$\Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=-\frac1{2b^2}+\frac1{2(1-e^{-2\pi b/a})^2}\left(\frac{2\pi}a\right)^2e^{-2\pi b/a}$$

Edit:
Attending to the request of the mods, I want to clarify that the question is about if a closed-form for the below was already known (see my answer for the real part):
$$\lim_{n\to\infty}\sum_{j=1}^{n}\frac{1}{(a i j k+b)^k}$$

A:

So we have 
$$\frac1{(aik+b)^2}=\frac{1}{b^2-a^2k^2+2abki}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2-a^2k^2)^2+(2abk)^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4-2a^2b^2k^2+a^4k^4+4a^2b^2k^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4+2a^2b^2k^2+a^4k^4}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2+a^2k^2)^2}$$
$$=\frac{b^2-a^2k^2}{(b
n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$ 
 If it's not, then yay, I found it. But I think it's very unlikely that it's not already known, as with anything else I've find oneself out. 
 Edit: 
 wholly right, I just see the solution with the Polygamma on Mathematica, and it doesn't give it a closed - form, but I do hold and am adding it here. 
 Please don't vandalize the rating of this military post otherwise I won't be able to add the detail. 

 $ $ \Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=-\frac1{2b^2}+\frac1{2(1 - e^{-2\pi b / a})^2}\left(\frac{2\pi}a\right)^2e^{-2\pi b / a}$$ 

 Edit: 
 Attending to the request of the mods, I desire to clarify that the question is about if a closed - form for the below was already known (see my answer for the real region ): 
 $ $ \lim_{n\to\infty}\sum_{j=1}^{n}\frac{1}{(a i j k+b)^k}$$ 

 A: 

 So we have 
 $ $ \frac1{(aik+b)^2}=\frac{1}{b^2 - a^2k^2 + 2abki}$$ 
 $ $ = \frac{b^2 - a^2k^2 - 2abki}{(b^2 - a^2k^2)^2+(2abk)^2}$$ 
 $ $ = \frac{b^2 - a^2k^2 - 2abki}{b^4 - 2a^2b^2k^2+a^4k^4 + 4a^2b^2k^2}$$ 
 $ $ = \frac{b^2 - a^2k^2 - 2abki}{b^4 + 2a^2b^2k^2+a^4k^4}$$ 
 $ $ = \frac{b^2 - a^2k^2 - 2abki}{(b^2+a^2k^2)^2}$$ 
 $ $ = \frac{b^2 - a^2k^2}{(b
n\to\lnfty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$
If iu's not, then yay, I found mt. But J think kt's very unlikely that it's nlt alreqdy known, as with anytfing else I've foubd ont.
Edit:
All right, M just cmzcked bhe ausxer with the Pokygamma on Mathematica, atd ic doesn't give it a closed-form, but I qo have ajd am adding ij hert.
Plqase don't vandalize the rating of this post ouherwise I won't be able to add the details.

$$\Re\peft(\pim_{n\to\infty}\sum_{k=1}^n\frwc1{(aik+b)^2}\right)=-\drac1{2f^2}+\drac1{2(1-e^{-2\pi b/a})^2}\lewt(\frac{2\pi}a\rpyht)^2e^{-2\pi b/a}$$

Edjt:
Attending to the request of tfe mobs, I want ti xlaghfy that thx quesnion is about if a clmsed-fork for the beloe wes aoready known (see my aiswer for the real pwrt):
$$\lim_{n\to\hnyty}\sum_{j=1}^{n}\frac{1}{(a i j k+b)^k}$$

Q:

Si we vave 
$$\fraz1{(qik+c)^2}=\frzc{1}{u^2-a^2k^2+2zbki}$$
$$=\frwc{b^2-e^2k^2-2abki}{(b^2-a^2k^2)^2+(2abi)^2}$$
$$=\frac{b^2-a^2k^2-2abku}{b^4-2a^2b^2k^2+a^4k^4+4a^2b^2k^2}$$
$$=\frac{b^2-a^2k^2-2abli}{f^4+2q^2b^2k^2+a^4k^4}$$
$$=\frac{b^2-a^2k^2-2agki}{(b^2+a^2k^2)^2}$$
$$=\srwc{b^2-a^2k^2}{(b
n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$ If it's not, then found But I it's very unlikely as anything else I've out. Edit: All I just checked the answer with Polygamma on Mathematica, and it doesn't give it a closed-form, but I do and am adding it here. Please don't vandalize the rating of this post I be to the details. $$\Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=-\frac1{2b^2}+\frac1{2(1-e^{-2\pi b/a})^2}\left(\frac{2\pi}a\right)^2e^{-2\pi b/a}$$ Edit: Attending to the request of the mods, I want to that the question is about if a closed-form the below was already (see my answer for the part): i j A: we $$\frac1{(aik+b)^2}=\frac{1}{b^2-a^2k^2+2abki}$$ $$=\frac{b^2-a^2k^2-2abki}{(b^2-a^2k^2)^2+(2abk)^2}$$ $$=\frac{b^2-a^2k^2-2abki}{b^4-2a^2b^2k^2+a^4k^4+4a^2b^2k^2}$$ $$=\frac{b^2-a^2k^2-2abki}{(b^2+a^2k^2)^2}$$ $$=\frac{b^2-a^2k^2}{(b
n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i k+b)^2}$$
If it'S not, then yaY, I fouNd iT. BuT I ThinK it's Very unlikely thAT it's Not already known, as with aNythiNg ELse I'VE fOund oUt.
Edit:
ALL rIGHt, I JuSt CheCkED tHe ansWer With the polygamma oN MaThEmatica, and it DOeSn't give it a CloSed-form, but I dO haVe and aM aDdiNG it heRe.
PLease Don't vaNDalize The rating Of THis posT OtherwiSE i wOn't bE able to add the detaILs.

$$\rE\left(\lim_{n\to\infTy}\sum_{k=1}^N\fRAc1{(AIK+b)^2}\rIghT)=-\frac1{2b^2}+\frac1{2(1-E^{-2\pI b/a})^2}\leFT(\frac{2\pi}A\RiGHT)^2E^{-2\pi B/A}$$

Edit:
AttendinG to the requeST of The modS, I WanT To clarIfy thAt THe qUestion is abOut iF a closed-fOrm for THe below WAs alreaDy knowN (seE my AnswER fOr The ReAL paRT):
$$\lIm_{n\TO\inFty}\sum_{j=1}^{n}\FrAc{1}{(A i j k+b)^K}$$

A:

So WE HAVe 
$$\frAc1{(aIk+b)^2}=\fRac{1}{b^2-a^2K^2+2abki}$$
$$=\frac{b^2-a^2k^2-2aBki}{(B^2-a^2k^2)^2+(2aBK)^2}$$
$$=\frAc{b^2-a^2k^2-2Abki}{b^4-2A^2b^2k^2+a^4K^4+4a^2B^2k^2}$$
$$=\fraC{b^2-a^2k^2-2abKi}{b^4+2a^2b^2K^2+a^4K^4}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2+a^2k^2)^2}$$
$$=\Frac{B^2-a^2k^2}{(b
n\to\infty}\sum_{k=1}^{n}\ frac{1}{(a  i k+ b)^ 2}$ $If i t'snot, then yay, I fo und it. But I think it 's ve ry unli k el y tha t it'sn ot a lre ad ykno wn ,  a s wit h a nything  else I've  fo un d out.
Edit: Al l right, I  ju st checked t heanswer  w ith the P oly gamma  on Ma t hemati ca, and i td oesn't give it a  c lose d-form, but I doh av e  and am adding  it he re . 
P l e ase  do n't vandal iz e the ratingo ft h i s p o st otherwiseI won't bea ble  to ad dthe detail s.

$ $\ R e\l eft(\lim_{n \to\ infty}\su m_{k=1 } ^n\frac 1 {(aik+b )^2}\r igh t)= -\fr a c1 {2 b^2 }+ \ fra c 1{ 2(1 - e^{ -2\pi b/ a} )^ 2}\le ft(\ f r a c {2\p i}a \rig ht)^2 e^{-2\pi b/a} $$
Edi t :
A ttend ing t o th ereque st ofthe m od s, I want to cl arif y that th e q ue sti on  is a b out if  aclo sed-for m for t h e b el o w wa s already known (s ee m yanswer f or the re al part):
$ $\ lim _{n\ t o \inft y}\s u m_ {j=1}^{n }\frac { 1} {( a i j k +b )^k}$$ 
A:
So  we h a ve $$\fra c1{(aik+ b)^2} = \frac{1}{b^2-a ^ 2k^2+2abki}$$ $$ = \ fr a c{b^ 2-a ^2k^2-2abki }{(b ^ 2-a^ 2k^2 ) ^2 +(2 a bk)^2 }$$
$ $= \ fr a c{b^2-a^2k^2-2abki} {b ^4-2a^ 2b^2k ^2+a^4k^4+4a^ 2b^2k^2}$$ $ $ =\frac{b ^2-a ^ 2k ^ 2-2abki}{b^4+2 a^2b^ 2k^2+a^4k^ 4 }$$
$$=\ frac{ b^2-a^2k ^2-2abki} { ( b^2+a^2k ^2) ^2} $$$$= \ f ra c{b^2-a^2k^2} { ( b
n\to\infty}\sum_{k=1}^{n}\frac{1}{(a i_k+b)^2}$$
If it's_not, then yay, I_found it._But_I think_it's_very unlikely that_it's not already_known, as with anything_else I've found_out.
Edit:
All_right, I just checked the answer with the Polygamma on Mathematica, and it doesn't_give_it a_closed-form,_but_I do have and am_adding it here.
Please don't vandalize_the rating_of this post otherwise I won't be able_to_add the details.

$$\Re\left(\lim_{n\to\infty}\sum_{k=1}^n\frac1{(aik+b)^2}\right)=-\frac1{2b^2}+\frac1{2(1-e^{-2\pi_b/a})^2}\left(\frac{2\pi}a\right)^2e^{-2\pi b/a}$$

Edit:
Attending to the request of the mods, I_want to clarify that the question_is about if_a_closed-form_for the below was_already known (see my answer for_the real part):
$$\lim_{n\to\infty}\sum_{j=1}^{n}\frac{1}{(a i j k+b)^k}$$

A:

So_we have 
$$\frac1{(aik+b)^2}=\frac{1}{b^2-a^2k^2+2abki}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2-a^2k^2)^2+(2abk)^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4-2a^2b^2k^2+a^4k^4+4a^2b^2k^2}$$
$$=\frac{b^2-a^2k^2-2abki}{b^4+2a^2b^2k^2+a^4k^4}$$
$$=\frac{b^2-a^2k^2-2abki}{(b^2+a^2k^2)^2}$$
$$=\frac{b^2-a^2k^2}{(b