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Sum of series:
5/n
1/(1*3)+1/((2n-1)*(2n+1))
1/((5n-4)(5n+1))
arcsin(n/(n^2+1))
Identical expressions
one /((5n- four)(5n+ one))
1 divide by ((5n minus 4)(5n plus 1))
one divide by ((5n minus four)(5n plus one))
1/5n-45n+1
1 divide by ((5n-4)(5n+1))
Similar expressions
1/((5n-4)(5n-1))
1/((5n+4)(5n+1))
(1)/5*n-4*5*n+1

Sum of series/ 1/((5n-4)(5n+1))

Sum of series 1/((5n-4)(5n+1))

The solution

You have entered [src]

  oo                     
 ___                     
 \  `                    
  \            1         
   )  -------------------
  /   (5*n - 4)*(5*n + 1)
 /__,                    
n = 1

\sum_{n=1}^{\infty} \frac{1}{\left(5 n - 4\right) \left(5 n + 1\right)}

Sum(1/((5*n - 4)*(5*n + 1)), (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{1}{\left(5 n - 4\right) \left(5 n + 1\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(5 n - 4\right) \left(5 n + 1\right)}

and

x_{0} = 0

d = 0

c = 1

then

1 = \lim_{n \to \infty}\left(\left(5 n + 6\right) \left|{\frac{1}{5 n - 4}}\right|\right)

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

Gamma(11/5) 
------------
6*Gamma(6/5)

\frac{\Gamma\left(\frac{11}{5}\right)}{6 \Gamma\left(\frac{6}{5}\right)}

gamma(11/5)/(6*gamma(6/5))

Numerical answer [src]

0.200000000000000000000000000000

0.200000000000000000000000000000

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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Identical expressions

Similar expressions

Sum of series 1/((5n-4)(5n+1))

The solution

Examples of finding the sum of a series