Mister Exam

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Sum of series:
cos(2*n)/2^n
(n-1)/(n(n+1)(n+2))
n^100*100^n/factorial(n)
1/(log(log(n))^(log(n)))
Identical expressions
(n- one)/(n(n+ one)(n+ two))
(n minus 1) divide by (n(n plus 1)(n plus 2))
(n minus one) divide by (n(n plus one)(n plus two))
n-1/nn+1n+2
(n-1) divide by (n(n+1)(n+2))
Similar expressions
(n-1)/(n(n-1)(n+2))
(n+1)/(n(n+1)(n+2))
n-1/(n(n+1)(n+2))
(n-1)/(n(n+1)(n-2))

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

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

The solution

You have entered [src]

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

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

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

The radius of convergence of the power series

Given number:

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

and

x_{0} = 0

d = 0

c = 1

then

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

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

1/4

\frac{1}{4}

1/4

Numerical answer [src]

0.250000000000000000000000000000

0.250000000000000000000000000000

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

The solution

Examples of finding the sum of a series