Mister Exam

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Sum of series:
1/n^2
(-1)n/2n-1
lnn/n^2
1/((2n-1)(2n)(2n+1))
Identical expressions
one /((2n- one)(2n)(2n+ one))
1 divide by ((2n minus 1)(2n)(2n plus 1))
one divide by ((2n minus one)(2n)(2n plus one))
1/2n-12n2n+1
1 divide by ((2n-1)(2n)(2n+1))
Similar expressions
1/((2n-1)(2n)(2n-1))
1/(2n-1)(2n)(2n+1)
1/((2n+1)(2n)(2n+1))

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

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

The solution

You have entered [src]

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

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

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

The radius of convergence of the power series

Given number:

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

and

x_{0} = 0

d = 0

c = 1

then

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

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

nan

\text{NaN}

nan

Numerical answer [src]

0.193147180559945309417232121458

0.193147180559945309417232121458

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

The solution

Examples of finding the sum of a series