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Sum of series:
1/((3n-1)(3n+2))
n/3^n
x*y*z
0.4^n
Identical expressions
one /((3n- one)(3n+ two))
1 divide by ((3n minus 1)(3n plus 2))
one divide by ((3n minus one)(3n plus two))
1/3n-13n+2
1 divide by ((3n-1)(3n+2))
Similar expressions
1/((3n-1)*(3n+2))
1/((3*n-1)*(3*n+2))
1/((3n+1)(3n+2))
1/((3*n-1)*(3n+2))
1/(3n-1)*(3n+2)
1/(3*n-1)*(3*n+2)
1/((3n-1)(3n-2))

Sum of series/ 1/((3n-1)(3n+2))

Sum of series 1/((3n-1)(3n+2))

The solution

You have entered [src]

  oo                     
 ___                     
 \  `                    
  \            1         
   )  -------------------
  /   (3*n - 1)*(3*n + 2)
 /__,                    
n = 1

\sum_{n=1}^{\infty} \frac{1}{\left(3 n - 1\right) \left(3 n + 2\right)}

Sum(1/((3*n - 1)*(3*n + 2)), (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{1}{\left(3 n - 1\right) \left(3 n + 2\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{1}{\left(3 n - 1\right) \left(3 n + 2\right)}

and

x_{0} = 0

d = 0

c = 1

then

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

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

  Gamma(8/3) 
-------------
10*Gamma(5/3)

\frac{\Gamma\left(\frac{8}{3}\right)}{10 \Gamma\left(\frac{5}{3}\right)}

gamma(8/3)/(10*gamma(5/3))

Numerical answer [src]

0.166666666666666666666666666667

0.166666666666666666666666666667

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 1/((3n-1)(3n+2))

The solution

Examples of finding the sum of a series