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Sum of series:
6/((3n-1)*(3n+5))
(-1)^(n+1)/n
1/((4k^2)-1)
7^(n+1)/(4^(n-2)*9^n)
Identical expressions
six /((3n- one)*(3n+ five))
6 divide by ((3n minus 1) multiply by (3n plus 5))
six divide by ((3n minus one) multiply by (3n plus five))
6/((3n-1)(3n+5))
6/3n-13n+5
6 divide by ((3n-1)*(3n+5))
Similar expressions
6/(3*n-1)*(3*n+5)
6/(3n-1*3n+5)
6/((3n+1)*(3n+5))
6/(3n-1)(3n+5)
6/((3*n-1)*(3*n+5))
6/((3n-1)(3n+5))
6/((3n-1)*(3n-5))

Sum of series/ 6/((3n-1)*(3n+5))

Sum of series 6/((3n-1)*(3n+5))

The solution

You have entered [src]

  oo                     
 ___                     
 \  `                    
  \            6         
   )  -------------------
  /   (3*n - 1)*(3*n + 5)
 /__,                    
n = 1

\sum_{n=1}^{\infty} \frac{6}{\left(3 n - 1\right) \left(3 n + 5\right)}

Sum(6/(((3*n - 1)*(3*n + 5))), (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{6}{\left(3 n - 1\right) \left(3 n + 5\right)}

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} = \frac{6}{\left(3 n - 1\right) \left(3 n + 5\right)}

and

x_{0} = 0

d = 0

c = 1

then

1 = \lim_{n \to \infty}\left(\frac{\left(3 n + 2\right) \left(3 n + 8\right) \left|{\frac{1}{3 n - 1}}\right|}{3 n + 5}\right)

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

21*Gamma(11/3)
--------------
80*Gamma(8/3)

\frac{21 \Gamma\left(\frac{11}{3}\right)}{80 \Gamma\left(\frac{8}{3}\right)}

21*gamma(11/3)/(80*gamma(8/3))

Numerical answer [src]

0.700000000000000000000000000000

0.700000000000000000000000000000

The graph

Examples of finding the sum of a series

The Sum of the Power Series
```
x^n/n
```
```
(x-1)^n
```
Factorial
```
1/2^(n!)
```
```
n^2/n!
```
```
x^n/n!
```
```
k!/(n!*(n+k)!)
```
Flint Hills Series
```
csc(n)^2/n^3
```
Basel problem
```
1/n^2
```
```
1/n^4
```
```
1/n^6
```
Harmonic series
```
1/n
```
Grandi's series
```
(-1)^n
```
Alternating series
```
(-1)^(n + 1)/n
```
```
(n + 2)*(-1)^(n - 1)
```
```
(3*n - 1)/(-5)^n
```
```
(-1)^(n - 1)*n/(6*n - 5)
```
Newton–Mercator series
```
(-1)^(n + 1)/n*x^n
```
Exam the series for convergence
```
(3*n - 1)/(-5)^n
```

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How to use it?

Identical expressions

Similar expressions

Sum of series 6/((3n-1)*(3n+5))

The solution

Examples of finding the sum of a series