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1/(10^(4n-1))

  • How to use it?

  • Sum of series:
  • 6/9n^2+12n-5 6/9n^2+12n-5
  • 5^n 5^n
  • n^n/2^(n+1) n^n/2^(n+1)
  • n^3 n^3

  • Identical expressions

  • one /(ten ^(4n- one))
  • 1 divide by (10 to the power of (4n minus 1))
  • one divide by (ten to the power of (4n minus one))
  • 1/(10(4n-1))
  • 1/104n-1
  • 1/10^4n-1
  • 1 divide by (10^(4n-1))

  • Similar expressions

  • 1/(10^(4n+1))
Sum of series/ 1/(10^(4n-1))

Sum of series 1/(10^(4n-1))


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The solution

You have entered [src] 
  oo           
____           
\   `          
 \        1    
  \   ---------
  /     4*n - 1
 /    10       
/___,          
n = 0
$$\sum_{n=0}^{\infty} \frac{1}{10^{4 n - 1}}$$ 
Sum(1/(10^(4*n - 1)), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{10^{4 n - 1}}$$
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
$$a_{n} = 10^{1 - 4 n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(10^{1 - 4 n} 10^{4 n + 3}\right)$$
Let's take the limit
we find
False

False
The rate of convergence of the power series
The answer [src] 
100000
------
 9999
$$\frac{100000}{9999}$$ 
100000/9999
Numerical answer [src] 
10.0010001000100010001000100010
10.0010001000100010001000100010
The graph
Sum of series 1/(10^(4n-1))

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