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n^2*sin(5/(3^n))

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  • Sum of series:
  • n^2*sin(5/(3^n)) n^2*sin(5/(3^n))
  • n*2^n n*2^n
  • n^3 n^3
  • (2n+1)/(n(n+2)(n+4)) (2n+1)/(n(n+2)(n+4))

  • Identical expressions

  • n^ two *sin(five /(three ^n))
  • n squared multiply by sinus of (5 divide by (3 to the power of n))
  • n to the power of two multiply by sinus of (five divide by (three to the power of n))
  • n2*sin(5/(3n))
  • n2*sin5/3n
  • n²*sin(5/(3^n))
  • n to the power of 2*sin(5/(3 to the power of n))
  • n^2sin(5/(3^n))
  • n2sin(5/(3n))
  • n2sin5/3n
  • n^2sin5/3^n
  • n^2*sin(5 divide by (3^n))
Sum of series/ n^2*sin(5/(3^n))

Sum of series n^2*sin(5/(3^n))


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The solution

You have entered [src] 
  oo            
____            
\   `           
 \     2    /5 \
  \   n *sin|--|
  /         | n|
 /          \3 /
/___,           
n = 1
$$\sum_{n=1}^{\infty} n^{2} \sin{\left(\frac{5}{3^{n}} \right)}$$ 
Sum(n^2*sin(5/3^n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$n^{2} \sin{\left(\frac{5}{3^{n}} \right)}$$
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
$$a_{n} = n^{2} \sin{\left(5 \cdot 3^{- n} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{n^{2} \left|{\frac{\sin{\left(5 \cdot 3^{- n} \right)}}{\sin{\left(5 \cdot 3^{- n - 1} \right)}}}\right|}{\left(n + 1\right)^{2}}\right)$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
The answer [src] 
  oo               
 ___               
 \  `              
  \    2    /   -n\
  /   n *sin\5*3  /
 /__,              
n = 1
$$\sum_{n=1}^{\infty} n^{2} \sin{\left(5 \cdot 3^{- n} \right)}$$ 
Sum(n^2*sin(5*3^(-n)), (n, 1, oo))
Numerical answer [src] 
6.70600550431566704771021841536
6.70600550431566704771021841536
The graph
Sum of series n^2*sin(5/(3^n))

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