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n^n/2^(n+1)

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  • Sum of series:
  • 6/9n^2+12n-5 6/9n^2+12n-5
  • n^n/2^(n+1) n^n/2^(n+1)
  • e^(-n) e^(-n)
  • (5^n-4^n)/20^n (5^n-4^n)/20^n

  • Identical expressions

  • n^n/ two ^(n+ one)
  • n to the power of n divide by 2 to the power of (n plus 1)
  • n to the power of n divide by two to the power of (n plus one)
  • nn/2(n+1)
  • nn/2n+1
  • n^n/2^n+1
  • n^n divide by 2^(n+1)

  • Similar expressions

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

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


∑
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The solution

You have entered [src] 
  oo        
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\   `       
 \       n  
  \     n   
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  /    n + 1
 /    2     
/___,       
n = 1
∑n=1∞nn2n+1\sum_{n=1}^{\infty} \frac{n^{n}}{2^{n + 1}} 
Sum(n^n/2^(n + 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
nn2n+1\frac{n^{n}}{2^{n + 1}}
It is a series of species
an(cx−x0)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+lim⁡n→∞∣anan+1∣cR^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}
In this case
an=2−n−1nna_{n} = 2^{- n - 1} n^{n}
and
x0=0x_{0} = 0
,
d=0d = 0
,
c=1c = 1
then
1=lim⁡n→∞(2−n−1⋅2n+2nn(n+1)−n−1)1 = \lim_{n \to \infty}\left(2^{- n - 1} \cdot 2^{n + 2} n^{n} \left(n + 1\right)^{- n - 1}\right)
Let's take the limit
we find
False

False
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.505000
The answer [src] 
  oo            
 ___            
 \  `           
  \    -1 - n  n
  /   2      *n 
 /__,           
n = 1
∑n=1∞2−n−1nn\sum_{n=1}^{\infty} 2^{- n - 1} n^{n} 
Sum(2^(-1 - n)*n^n, (n, 1, oo))
The graph
Sum of series n^n/2^(n+1)

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