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

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  • Sum of series:
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  • sin1/n sin1/n
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  • (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)

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  • (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
∑n=1∞3n+15n\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:
3n+15n\frac{3 n + 1}{5^{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=3n+1a_{n} = 3 n + 1
and
x0=−5x_{0} = -5
,
d=−1d = -1
,
c=0c = 0
then
1R=∞~(−5+lim⁡n→∞(3n+13n+4))\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=0R = 0
The rate of convergence of the power series
1.07.01.52.02.53.03.54.04.55.05.56.06.50.51.5
The answer [src] 
19
--
16
1916\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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