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(2*n-1)/2^n

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  • Sum of series:
  • (2*n-1)/2^n (2*n-1)/2^n
  • (-1)^n (-1)^n
  • 6/(9n^2+6n-8) 6/(9n^2+6n-8)
  • x^(3n)/n!

  • Identical expressions

  • (two *n- one)/ two ^n
  • (2 multiply by n minus 1) divide by 2 to the power of n
  • (two multiply by n minus one) divide by two to the power of n
  • (2*n-1)/2n
  • 2*n-1/2n
  • (2n-1)/2^n
  • (2n-1)/2n
  • 2n-1/2n
  • 2n-1/2^n
  • (2*n-1) divide by 2^n

  • Similar expressions

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

Sum of series (2*n-1)/2^n


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \    2*n - 1
  \   -------
  /       n  
 /       2   
/___,        
n = 1
∑n=1∞2n−12n\sum_{n=1}^{\infty} \frac{2 n - 1}{2^{n}} 
Sum((2*n - 1)/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
2n−12n\frac{2 n - 1}{2^{n}}
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=2n−1a_{n} = 2 n - 1
and
x0=−2x_{0} = -2
,
d=−1d = -1
,
c=0c = 0
then
1R=∞~(−2+lim⁡n→∞(∣2n−1∣2n+1))\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty}\left(\frac{\left|{2 n - 1}\right|}{2 n + 1}\right)\right)
Let's take the limit
we find
False

R=0R = 0
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.05.0
The answer [src] 
3
33 
3
Numerical answer [src] 
3.00000000000000000000000000000
3.00000000000000000000000000000
The graph
Sum of series (2*n-1)/2^n

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