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Sum of series:
1/(n(n+1))
12/(n^2+4*n+3)
(1-5/n)^n
i^n
Identical expressions
twelve /(n^ two + four *n+ three)
12 divide by (n squared plus 4 multiply by n plus 3)
twelve divide by (n to the power of two plus four multiply by n plus three)
12/(n2+4*n+3)
12/n2+4*n+3
12/(n²+4*n+3)
12/(n to the power of 2+4*n+3)
12/(n^2+4n+3)
12/(n2+4n+3)
12/n2+4n+3
12/n^2+4n+3
12 divide by (n^2+4*n+3)
Similar expressions
12/(n^2-4*n+3)
12/(n^2+4*n-3)

Sum of series/ 12/(n^2+4*n+3)

Sum of series 12/(n^2+4*n+3)

The solution

You have entered [src]

  oo              
____              
\   `             
 \         12     
  \   ------------
  /    2          
 /    n  + 4*n + 3
/___,             
n = 1

\sum_{n=1}^{\infty} \frac{12}{\left(n^{2} + 4 n\right) + 3}

Sum(12/(n^2 + 4*n + 3), (n, 1, oo))

The radius of convergence of the power series

Given number:

\frac{12}{\left(n^{2} + 4 n\right) + 3}

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{12}{n^{2} + 4 n + 3}

and

x_{0} = 0

d = 0

c = 1

then

1 = \lim_{n \to \infty}\left(\frac{12 \left(\frac{n}{3} + \frac{\left(n + 1\right)^{2}}{12} + \frac{7}{12}\right)}{n^{2} + 4 n + 3}\right)

Let's take the limit
we find

True

False

The rate of convergence of the power series

The answer [src]

5

Numerical answer [src]

5.00000000000000000000000000000

5.00000000000000000000000000000

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 12/(n^2+4*n+3)

The solution

Examples of finding the sum of a series