Apply the Ratio Test to the given series by considering the limit: lim n → ∞ | a n + 1 a n | | x | , {\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right||x|,} where a n = 3 n + n 3 {\displaystyle a_{n}=3^{n}+n^{3}} . Then we have
lim n → ∞ 3 n + 1 + ( n + 1 ) 3 3 n + n 3 | x | = lim n → ∞ 3 + ( n + 1 ) 3 / 3 n 1 + n 3 / 3 n | x | = 3 | x | . {\displaystyle \lim _{n\to \infty }{\frac {3^{n+1}+(n+1)^{3}}{3^{n}+n^{3}}}|x|=\lim _{n\to \infty }{\frac {3+(n+1)^{3}/3^{n}}{1+n^{3}/3^{n}}}|x|=3|x|.}
For convergence of the power series we want
3 | x | < 1 , i . e . , | x | < 1 / 3. {\displaystyle 3|x|<1,\quad \mathrm {i.e.,} \quad |x|<1/3.}
Thus, ratio of convergence is given by 1 / 3 {\displaystyle 1/3} .
where 
. Then we have
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