(\frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\right\}$$
$$\sqrt[n]{1+2^n+(-3)^n}=\sqrt[n]{(-3)^n\left(\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1}$$
I tried to manipulate with the above expression and apply the squeeze theorem, but I haven't managed to get anything useful for this problem, except:
$$\sqrt[n]{1+2^n+(-3)^n}<\sqrt[n]{3\cdot3^n}=3\sqrt[n]{3}$$ I also considered artithmetic inequalities and the comparison: $1+2^n\leq(1+2)^n$, unsuccessfully.
I would definitely examine the cases $n\pmod{6}$ or $n\pmod{3}$. How can I move further from this point and what can I do with the $n-\text{th}$ root? I thought I could see how one of the factors behaves since $\cos\left(\frac{2\pi n}{3}\right)$ alternates for $n\in\mathbb N$. Thank you in advance!

A:

When $n$ is even, 
$$\tag1
(1+2^n+(-3)^n)^{1/n}=(1+2^n+3^n)^{1/n}\to3.
$$
when $n$ is odd, 
$$
(1+2^n+(-3)^n)^{1/n}\to-3.
$$
If you consider any subsequence $a_{n_k}$ that contains both even and odd $n_k$ for arbitrarily big $k$, the subsequence will fail to be convergent. So we only get convergent subsequences when the $n_k$ are eventually all even or odd. Thus the possible limits are 
$$
-\tfrac12\times3,-\tfrac12\times(-3),3,-3.
$$
That is, the accumulation points are 
$$
-\frac32,\frac32,3,-3. 
$$

To calculate the
( \frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\right\}$$ 
 $ $ \sqrt[n]{1 + 2^n+(-3)^n}=\sqrt[n]{(-3)^n\left(\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1}$$ 
 I tried to manipulate with the above expression and apply the hug theorem, but I haven't cope to get anything useful for this problem, except: 
 $ $ \sqrt[n]{1 + 2^n+(-3)^n}<\sqrt[n]{3\cdot3^n}=3\sqrt[n]{3}$$ I besides considered artithmetic inequalities and the comparison: $ 1 + 2^n\leq(1 + 2)^n$, unsuccessfully. 
 I would definitely analyze the cases $ n\pmod{6}$ or $ n\pmod{3}$. How can I move farther from this point and what can I do with the $ n-\text{th}$ etymon? I thought I could see how one of the factor behave since $ \cos\left(\frac{2\pi n}{3}\right)$ alternates for $ n\in\mathbb N$. Thank you in advance! 

 A: 

 When $ n$ is even, 
 $ $ \tag1 
 (1 + 2^n+(-3)^n)^{1 / n}=(1 + 2^n+3^n)^{1 / n}\to3. 
 $ $ 
 when $ n$ is odd, 
 $ $ 
 (1 + 2^n+(-3)^n)^{1 / n}\to-3. 
 $ $ 
 If you consider any sequel $ a_{n_k}$ that contains both even and odd $ n_k$ for arbitrarily adult $ k$, the subsequence will fail to be convergent. So we only get convergent sequel when the $ n_k$ are eventually all even or odd. Thus the possible limits are 
 $ $ 
 -\tfrac12\times3,-\tfrac12\times(-3),3,-3. 
 $ $ 
 That is, the accretion points are 
 $ $ 
 -\frac32,\frac32,3,-3. 
 $ $ 

 To calculate the
(\fraf{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\righu\}$$
$$\sqrt[n]{1+2^n+(-3)^n}=\sqrt[n]{(-3)^n\lefj(\lwft(-\frar{1}{3}\right)^n+\meft(-\frac{2}{3}\fight)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\righv)^n+\ledt(-\frax{2}{3}\right)^n+1}$$
I tried to manioulate winh the abive txpression and ap'my the squeezs themcem, but I haven'j managed to get anything gsdfbl for this problem, except:
$$\sqrt[n]{1+2^n+(-3)^n}<\sqre[n]{3\cdot3^n}=3\xqgt[n]{3}$$ I also confidegeq arfptmmetic inequalities and the compzrison: $1+2^i\leq(1+2)^n$, unsuccessgully.
I would definitely exwminf the cases $n\pmod{6}$ lr $n\pmod{3}$. Hoq cag I move furtfer from this point ans what can I do with the $n-\text{tf}$ rooc? I thought I coknd see how ine os the factors behavev since $\cos\left(\frac{2\pi n}{3}\rmght)$ alternates for $n\in\mavhbb N$. Thank you in wdvance!

A:

Wveu $n$ is even, 
$$\tag1
(1+2^n+(-3)^n)^{1/n}=(1+2^n+3^n)^{1/n}\ro3.
$$
qhen $t$ is odd, 
$$
(1+2^b+(-3)^n)^{1/n}\go-3.
$$
Ir bou consifer any subsesuence $a_{n_k}$ rhat contains both tveg and odd $n_k$ fkr arbytwarily big $k$, the subsequence will fail uo be convergent. So we only tet convergent subseqoences wheg the $n_k$ are eventually all even or odd. Thus the [ossiuld lnnits xee 
$$
-\tfrac12\times3,-\tfrac12\times(-3),3,-3.
$$
That is, the accumulation loonns are 
$$
-\frac32,\frac32,3,-3. 
$$

Tj calculate tje
(\frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\right\}$$ $$\sqrt[n]{1+2^n+(-3)^n}=\sqrt[n]{(-3)^n\left(\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1}$$ I tried to manipulate above and apply squeeze theorem, but anything for this problem, $$\sqrt[n]{1+2^n+(-3)^n}<\sqrt[n]{3\cdot3^n}=3\sqrt[n]{3}$$ I also artithmetic inequalities and the comparison: $1+2^n\leq(1+2)^n$, I would definitely examine the cases $n\pmod{6}$ or $n\pmod{3}$. How can I move from this point and what can I do with the $n-\text{th}$ root? I I see one the factors behaves since $\cos\left(\frac{2\pi n}{3}\right)$ alternates for $n\in\mathbb N$. Thank you in advance! A: When is even, $$\tag1 (1+2^n+(-3)^n)^{1/n}=(1+2^n+3^n)^{1/n}\to3. $$ when $n$ is $$ (1+2^n+(-3)^n)^{1/n}\to-3. $$ If consider any subsequence $a_{n_k}$ that both and odd for big the subsequence will to be convergent. So we only get convergent subsequences when the $n_k$ are eventually all even or Thus the are $$ $$ is, accumulation points are $$ To calculate the
(\frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\rigHt\}$$
$$\sqrt[n]{1+2^n+(-3)^n}=\sQrt[n]{(-3)^n\LefT(\leFt(-\Frac{1}{3}\RighT)^n+\left(-\frac{2}{3}\righT)^N+1\rigHt)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\lEft(-\frAc{2}{3}\RIght)^N+1}$$
i tRied tO manipuLAtE WIth ThE aBovE eXPrEssioN anD apply tHe squeeze tHeoReM, but I haven't mANaGed to get anYthIng useful for ThiS problEm, ExcEPt:
$$\sqrT[n]{1+2^n+(-3)^N}<\sqrt[N]{3\cdot3^n}=3\SQrt[n]{3}$$ I aLso considErED artitHMetic inEQUaLitiEs and the comparisoN: $1+2^N\lEQ(1+2)^n$, unsuccessfulLy.
I wouLd DEfINIteLy eXamine the cAsEs $n\pmOD{6}$ or $n\pmoD{3}$. hoW CAN I mOVe further from This point anD WhaT can I dO wIth THe $n-\texT{th}$ roOt? i ThoUght I could sEe hoW one of the FactorS Behaves SInce $\cos\Left(\frAc{2\pI n}{3}\rIght)$ ALtErNatEs FOr $n\IN\mAthBB N$. THank you iN aDvAnce!

A:

when $N$ IS EVen, 
$$\tAg1
(1+2^n+(-3)^N)^{1/n}=(1+2^n+3^n)^{1/N}\to3.
$$
whEn $n$ is odd, 
$$
(1+2^n+(-3)^n)^{1/n}\to-3.
$$
if yOu coNSidEr any SubseQuenCe $A_{n_k}$ thAt contAins bOtH even and odd $n_k$ foR arbItrarily bIg $k$, ThE suBsEquenCE will fAil To bE converGent. So wE OnlY gET COnVergent subsequenceS wHEN tHe $n_k$ are eVentuaLLy AlL Even or odD. THus The pOSSible LimiTS aRe 
$$
-\tfrac12\tImes3,-\tfRAc12\TiMes(-3),3,-3.
$$
That Is, The accUmUlaTioN poinTS are 
$$
-\Frac32,\frAc32,3,-3. 
$$

To calcUlate THe
(\frac{2\pi n}{3}\right)\i n\left\{-\ frac{ 1}{ 2}, 1\ righ t\}$ $
$$\sqrt[n]{1 + 2^n+ (-3)^n}=\sqrt[n]{(-3)^ n\lef t( \ left ( -\ frac{ 1}{3}\r i gh t ) ^n+ \l ef t(- \f r ac {2}{3 }\r ight)^n +1\right)} =-3 \s qrt[n]{\left ( -\ frac{1}{3} \ri ght)^n+\left (-\ frac{2 }{ 3}\ r ight) ^n+ 1}$$I trie d  to ma nipulatewi t h thea bove ex p r es sion  and apply the sq u ee z e theorem, but  I hav en ' tm a nag edto get any th ing u s eful fo r  t h i s  pr o blem, except: 
$$\sqrt[n] { 1+2 ^n+(-3 )^ n}< \ sqrt[n ]{3\c do t 3^n }=3\sqrt[n] {3}$ $ I alsoconsid e red art i thmetic  inequ ali tie s an d  t he  co mp a ris o n:  $1 + 2^n \leq(1+2 )^ n$ , uns ucce s s f u lly. 
Iwoul d def initely exami nethec ase s $n\ pmod{ 6}$or  $n\p mod{3} $. Ho wcan I move furt herfrom this  po in t a nd  what can Idowit h the $ n-\text { th} $r o o t?  I thought I could  s e e  h ow one o f thef ac to r s behave ssin ce $ \ c os\le ft(\ f ra c{2\pi n }{3}\r i gh t) $ alter na tes fo r$n\ in\ mathb b  N$.  Thank  you inadvan c e!

A:

When $ n $ is even, 
$ $ \t a g 1( 1+2^ n+( -3)^n)^{1/n }=(1 + 2^n+ 3^n) ^ {1 /n} \ to3.$$
wh en $n $  is odd, 
$$
(1+2^n +( -3)^n) ^{1/n }\to-3.
$$
If  you consi d e r  any sub sequ e nc e  $a_{n_k}$ tha t con tains both even and  odd$n_k$ fo r arbitra r i ly big $ k$,  th e s ubs e q ue nce will fail t o be  c onverge nt.  So weonl y g etcon ve rgent sub sequence swh en  t he$n_k$ are even tu all yall  even or odd . Thu s th epo s sib le limi t sa r e 
$ $-\ tfra c12 \t imes3 ,-\t f rac 12\time s(-3),3,- 3.$ $
Th at  i s, theaccumulationpo ints are $$ 
-\ frac32 , \ frac32,3 ,-3. 
$$

To calculatet he
(\frac{2\pi n}{3}\right)\in\left\{-\frac{1}{2},1\right\}$$
$$\sqrt[n]{1+2^n+(-3)^n}=\sqrt[n]{(-3)^n\left(\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1\right)}=-3\sqrt[n]{\left(-\frac{1}{3}\right)^n+\left(-\frac{2}{3}\right)^n+1}$$
I_tried to_manipulate with the above_expression and_apply_the squeeze_theorem,_but I haven't_managed to get_anything useful for this_problem, except:
$$\sqrt[n]{1+2^n+(-3)^n}<\sqrt[n]{3\cdot3^n}=3\sqrt[n]{3}$$ I_also_considered artithmetic inequalities and the comparison: $1+2^n\leq(1+2)^n$, unsuccessfully.
I would definitely examine the cases $n\pmod{6}$_or_$n\pmod{3}$. How_can_I_move further from this point_and what can I do_with the_$n-\text{th}$ root? I thought I could see how_one_of the factors_behaves since $\cos\left(\frac{2\pi n}{3}\right)$ alternates for $n\in\mathbb N$. Thank_you in advance!

A:

When $n$ is even,_
$$\tag1
(1+2^n+(-3)^n)^{1/n}=(1+2^n+3^n)^{1/n}\to3.
$$
when $n$ is_odd,_
$$
(1+2^n+(-3)^n)^{1/n}\to-3.
$$
If_you consider any subsequence_$a_{n_k}$ that contains both even and_odd $n_k$ for arbitrarily big $k$,_the subsequence will fail to be convergent._So we only get convergent subsequences_when the $n_k$ are eventually_all even_or odd. Thus the possible_limits are 
$$
-\tfrac12\times3,-\tfrac12\times(-3),3,-3.
$$
That_is, the_accumulation points are_
$$
-\frac32,\frac32,3,-3. 
$$

To calculate the