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(n^2+5)/2^n

  • How to use it?

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

  • Identical expressions

  • (n^ two + five)/ two ^n
  • (n squared plus 5) divide by 2 to the power of n
  • (n to the power of two plus five) divide by two to the power of n
  • (n2+5)/2n
  • n2+5/2n
  • (n²+5)/2^n
  • (n to the power of 2+5)/2 to the power of n
  • n^2+5/2^n
  • (n^2+5) divide by 2^n

  • Similar expressions

  • (n^2-5)/2^n
Sum of series/ (n^2+5)/2^n

Sum of series (n^2+5)/2^n


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The solution

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

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
11
$$11$$ 
11
Numerical answer [src] 
11.0000000000000000000000000000
11.0000000000000000000000000000
The graph
Sum of series (n^2+5)/2^n

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