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tan(1/2^n)/2^n

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  • Sum of series:
  • (1-5/n)^n (1-5/n)^n
  • 1/((2p)^2(2p-1)!(n-2p)!)
  • ((125/9))((3/5)^n) ((125/9))((3/5)^n)
  • tan(1/2^n)/2^n tan(1/2^n)/2^n

  • Identical expressions

  • tan(one / two ^n)/ two ^n
  • tangent of (1 divide by 2 to the power of n) divide by 2 to the power of n
  • tangent of (one divide by two to the power of n) divide by two to the power of n
  • tan(1/2n)/2n
  • tan1/2n/2n
  • tan1/2^n/2^n
  • tan(1 divide by 2^n) divide by 2^n
Sum of series/ tan(1/2^n)/2^n

Sum of series tan(1/2^n)/2^n


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \       / -n\
  \   tan\2  /
   )  --------
  /       n   
 /       2    
/___,         
n = 1
$$\sum_{n=1}^{\infty} \frac{\tan{\left(\left(\frac{1}{2}\right)^{n} \right)}}{2^{n}}$$ 
Sum(tan((1/2)^n)/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\tan{\left(\left(\frac{1}{2}\right)^{n} \right)}}{2^{n}}$$
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
$$a_{n} = \tan{\left(\left(\frac{1}{2}\right)^{n} \right)}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} \left|{\frac{\tan{\left(\left(\frac{1}{2}\right)^{n} \right)}}{\tan{\left(\left(\frac{1}{2}\right)^{n + 1} \right)}}}\right|\right)$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
The answer [src] 
  oo              
 ___              
 \  `             
  \    -n    / -n\
  /   2  *tan\2  /
 /__,             
n = 1
$$\sum_{n=1}^{\infty} 2^{- n} \tan{\left(\left(\frac{1}{2}\right)^{n} \right)}$$ 
Sum(2^(-n)*tan((1/2)^n), (n, 1, oo))
Numerical answer [src] 
0.357907384065669296993580013406
0.357907384065669296993580013406
The graph
Sum of series tan(1/2^n)/2^n

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