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(3*n+1)/(5^n)

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  • Sum of series:
  • n^2/3^n n^2/3^n
  • nx^n/(8n-9)*3^n
  • cos(n*x)/(n^2+1)
  • cos(i*n)/2^n cos(i*n)/2^n

  • Identical expressions

  • (three *n+ one)/(five ^n)
  • (3 multiply by n plus 1) divide by (5 to the power of n)
  • (three multiply by n plus one) divide by (five to the power of n)
  • (3*n+1)/(5n)
  • 3*n+1/5n
  • (3n+1)/(5^n)
  • (3n+1)/(5n)
  • 3n+1/5n
  • 3n+1/5^n
  • (3*n+1) divide by (5^n)

  • Similar expressions

  • (3*n-1)/(5^n)
Sum of series/ (3*n+1)/(5^n)

Sum of series (3*n+1)/(5^n)


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The solution

You have entered [src] 
  oo         
____         
\   `        
 \    3*n + 1
  \   -------
  /       n  
 /       5   
/___,        
n = 1
$$\sum_{n=1}^{\infty} \frac{3 n + 1}{5^{n}}$$ 
Sum((3*n + 1)/5^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3 n + 1}{5^{n}}$$
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
$$a_{n} = 3 n + 1$$
and
$$x_{0} = -5$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-5 + \lim_{n \to \infty}\left(\frac{3 n + 1}{3 n + 4}\right)\right)$$
Let's take the limit
we find
False

$$R = 0$$
The rate of convergence of the power series
The answer [src] 
19
--
16
$$\frac{19}{16}$$ 
19/16
Numerical answer [src] 
1.18750000000000000000000000000
1.18750000000000000000000000000
The graph
Sum of series (3*n+1)/(5^n)

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