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1/(4n^2+9)

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

  • Identical expressions

  • one /(4n^ two + nine)
  • 1 divide by (4n squared plus 9)
  • one divide by (4n to the power of two plus nine)
  • 1/(4n2+9)
  • 1/4n2+9
  • 1/(4n²+9)
  • 1/(4n to the power of 2+9)
  • 1/4n^2+9
  • 1 divide by (4n^2+9)

  • Similar expressions

  • 1/(4n^2-9)
Sum of series/ 1/(4n^2+9)

Sum of series 1/(4n^2+9)


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The solution

You have entered [src] 
  oo          
____          
\   `         
 \       1    
  \   --------
  /      2    
 /    4*n  + 9
/___,         
n = 1
$$\sum_{n=1}^{\infty} \frac{1}{4 n^{2} + 9}$$ 
Sum(1/(4*n^2 + 9), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{4 n^{2} + 9}$$
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
$$a_{n} = \frac{1}{4 n^{2} + 9}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{4 \left(n + 1\right)^{2} + 9}{4 n^{2} + 9}\right)$$
Let's take the limit
we find
True

False
The rate of convergence of the power series
Numerical answer [src] 
0.206286089822351285289635990323
0.206286089822351285289635990323
The graph
Sum of series 1/(4n^2+9)

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