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cos(n*100)/(n^2+1)

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  • Sum of series:
  • sin(1/n^2) sin(1/n^2)
  • factorial(x-n)
  • (n-1)*5 (n-1)*5
  • cosn/lnn cosn/lnn

  • Identical expressions

  • cos(n* one hundred)/(n^ two + one)
  • co sinus of e of (n multiply by 100) divide by (n squared plus 1)
  • co sinus of e of (n multiply by one hundred) divide by (n to the power of two plus one)
  • cos(n*100)/(n2+1)
  • cosn*100/n2+1
  • cos(n*100)/(n²+1)
  • cos(n*100)/(n to the power of 2+1)
  • cos(n100)/(n^2+1)
  • cos(n100)/(n2+1)
  • cosn100/n2+1
  • cosn100/n^2+1
  • cos(n*100) divide by (n^2+1)

  • Similar expressions

  • cos(n*100)/(n^2-1)
Sum of series/ cos(n*100)/(n^2+1)

Sum of series cos(n*100)/(n^2+1)


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The solution

You have entered [src] 
  oo            
____            
\   `           
 \    cos(n*100)
  \   ----------
  /      2      
 /      n  + 1  
/___,           
n = 1
$$\sum_{n=1}^{\infty} \frac{\cos{\left(100 n \right)}}{n^{2} + 1}$$ 
Sum(cos(n*100)/(n^2 + 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\cos{\left(100 n \right)}}{n^{2} + 1}$$
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
$$a_{n} = \frac{\cos{\left(100 n \right)}}{n^{2} + 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(\left(n + 1\right)^{2} + 1\right) \left|{\frac{\cos{\left(100 n \right)}}{\cos{\left(100 n + 100 \right)}}}\right|}{n^{2} + 1}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
The graph
Sum of series cos(n*100)/(n^2+1)

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