Find the radius of the series $$\sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}{(5+\cos(n\pi/3))^{n}}}$$
What is the radius of the convergence of the series? 
Please show clearly and help me how to solve this. Thank you! 
I know the $R=\dfrac{1}{\limsup|a_n|^{1/n}}$ but I cannot find value of limsup. My actual question to you is this!!?
 A: So let's try to apply the limsup formula for the radius of convergence, since you say this is your actual question.
First compute
$$
\sqrt[n]{|a_n|}=\frac{1}{n^{2/n}(5+\cos(n\pi/3))}.
$$
Recall that $\limsup x_n$ is the largest $x$ such that there exists a subsequence of $x_n$ converging to $x$. 
Note that $n^{2/n}=\exp
(2\log n/n)$ converges to $1$, so that 
$$
\limsup \sqrt[n]{|a_n|}=\limsup \frac{1}{5+\cos(n\pi/3)}.
$$
First oberve that 
$$
\frac{1}{5+\cos(n\pi/3)}\leq \frac{1}{5-1}=\frac{1}{4}
$$
for all $n$. So $\limsup \sqrt[n]{|a_n|}\leq 1/4$. 
And now for the extraction $n_k=3(2k+1)$, we have
$$
\frac{1}{5+\cos(n_k\pi/3)}=\frac{1}{4}.
$$
Hence $\limsup \sqrt[n]{|a_n|}\geq 1/4$.
Finally $\limsup \sqrt[n]{|a_n|}= 1/4$ and by the formula for the radius of convergence, $R=4$.
A: Edit: OP has changed the question, replacing $n^2$ with $n^{56}$, and $5$ with $23$, and $n\pi/3$ with $n\pi/7$. Fundamentally, nothing changes, the radius of convergence is now $22$, same argument. 
Original Question: This asked for the radius of convergence of $\displaystyle \sum_1^\infty \frac{x^n}{n^2(5+\sin(n\pi/3))^n}$.
Answer: Applying the Ratio Test, or the Root Test, shows that the radius of convergence is $\ge 4$. 
You may want to compare with the simpler series $\displaystyle\sum_1^\infty \frac{x^n}{n^24^n}$. 
To show that nothing $\gt 4$ will do, show that if $|x|\gt 4$, then the terms do not have limit $0$.
The issue is that $\cos(n\pi/3)$ is $-1$ for infinitely many $n$.
