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-(16/2^n)*n

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  • Sum of series:
  • 9/(9n^2+21n-8) 9/(9n^2+21n-8)
  • (3-sin*n)/n-lnn (3-sin*n)/n-lnn
  • 1/(1+n^2) 1/(1+n^2)
  • e^(-n) e^(-n)

  • Identical expressions

  • -(sixteen / two ^n)*n
  • minus (16 divide by 2 to the power of n) multiply by n
  • minus (sixteen divide by two to the power of n) multiply by n
  • -(16/2n)*n
  • -16/2n*n
  • -(16/2^n)n
  • -(16/2n)n
  • -16/2nn
  • -16/2^nn
  • -(16 divide by 2^n)*n

  • Similar expressions

  • (16/2^n)*n
Sum of series/ -(16/2^n)*n

Sum of series -(16/2^n)*n


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The solution

You have entered [src] 
  oo       
 ___       
 \  `      
  \     n  
  /   -8 *n
 /__,      
n = 1
$$\sum_{n=1}^{\infty} - 8^{n} n$$ 
Sum((-8^n)*n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$- 8^{n} n$$
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
$$a_{n} = - n$$
and
$$x_{0} = -8$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(-8 + \lim_{n \to \infty}\left(\frac{n}{n + 1}\right)\right)$$
Let's take the limit
we find
False
The rate of convergence of the power series
The answer [src] 
-oo
$$-\infty$$ 
-oo
Numerical answer
The series diverges
The graph
Sum of series -(16/2^n)*n

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