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sin(3n)/(7n)^(1/5)

Sum of series sin(3n)/(7n)^(1/5)



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The solution

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  oo          
____          
\   `         
 \    sin(3*n)
  \   --------
  /   5 _____ 
 /    \/ 7*n  
/___,         
n = 1         
n=1sin(3n)7n5\sum_{n=1}^{\infty} \frac{\sin{\left(3 n \right)}}{\sqrt[5]{7 n}}
Sum(sin(3*n)/(7*n)^(1/5), (n, 1, oo))
The radius of convergence of the power series
Given number:
sin(3n)7n5\frac{\sin{\left(3 n \right)}}{\sqrt[5]{7 n}}
It is a series of species
an(cxx0)dna_{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:
Rd=x0+limnanan+1cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=745sin(3n)7n5a_{n} = \frac{7^{\frac{4}{5}} \sin{\left(3 n \right)}}{7 \sqrt[5]{n}}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=limn(n+15sin(3n)sin(3n+3)n5)1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)
Let's take the limit
we find
1=limn(n+15sin(3n)sin(3n+3)n5)1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)
False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.5-0.5
The answer [src]
  oo               
____               
\   `              
 \     4/5         
  \   7   *sin(3*n)
   )  -------------
  /        5 ___   
 /       7*\/ n    
/___,              
n = 1              
n=1745sin(3n)7n5\sum_{n=1}^{\infty} \frac{7^{\frac{4}{5}} \sin{\left(3 n \right)}}{7 \sqrt[5]{n}}
Sum(7^(4/5)*sin(3*n)/(7*n^(1/5)), (n, 1, oo))
The graph
Sum of series sin(3n)/(7n)^(1/5)

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