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(sin(2^n))^2/n^2

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Sum of series/ (sin(2^n))^2/n^2

Sum of series (sin(2^n))^2/n^2


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \       2/ n\
  \   sin \2 /
   )  --------
  /       2   
 /       n    
/___,         
n = 1
$$\sum_{n=1}^{\infty} \frac{\sin^{2}{\left(2^{n} \right)}}{n^{2}}$$ 
Sum(sin(2^n)^2/n^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin^{2}{\left(2^{n} \right)}}{n^{2}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \frac{\sin^{2}{\left(2^{n} \right)}}{n^{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2} \sin^{2}{\left(2^{n} \right)} \left|{\frac{1}{\sin^{2}{\left(2^{n + 1} \right)}}}\right|}{n^{2}}\right)$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2} \sin^{2}{\left(2^{n} \right)} \left|{\frac{1}{\sin^{2}{\left(2^{n + 1} \right)}}}\right|}{n^{2}}\right)$$
False
The rate of convergence of the power series
The graph
Sum of series (sin(2^n))^2/n^2

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