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sqrt(n+1)/2^n

  • How to use it?

  • Sum of series:
  • 1/n(n+1) 1/n(n+1)
  • (3/7)^n (3/7)^n
  • 1/((3*n-2)*(3*n+1)) 1/((3*n-2)*(3*n+1))
  • sqrt(n+1)/2^n sqrt(n+1)/2^n

  • Identical expressions

  • sqrt(n+ one)/ two ^n
  • square root of (n plus 1) divide by 2 to the power of n
  • square root of (n plus one) divide by two to the power of n
  • √(n+1)/2^n
  • sqrt(n+1)/2n
  • sqrtn+1/2n
  • sqrtn+1/2^n
  • sqrt(n+1) divide by 2^n

  • Similar expressions

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

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


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The solution

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

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
  oo               
 ___               
 \  `              
  \    -n   _______
  /   2  *\/ 1 + n 
 /__,              
n = 1
$$\sum_{n=1}^{\infty} 2^{- n} \sqrt{n + 1}$$ 
Sum(2^(-n)*sqrt(1 + n), (n, 1, oo))
Numerical answer [src] 
1.69450750547150138439924692807
1.69450750547150138439924692807
The graph
Sum of series sqrt(n+1)/2^n

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