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factorial(2*n)/4^n

How to use it?
Sum of series:
|(-1)^n/3^n|
sqrt(n^2+3n)-sqrt(n^2-n)
sqrt(n)/(n+3)
factorial(2*n)/4^n
Identical expressions
factorial(two *n)/ four ^n
factorial(2 multiply by n) divide by 4 to the power of n
factorial(two multiply by n) divide by four to the power of n
factorial(2*n)/4n
factorial2*n/4n
factorial(2n)/4^n
factorial(2n)/4n
factorial2n/4n
factorial2n/4^n
factorial(2*n) divide by 4^n

Sum of series/ factorial(2*n)/4^n

Sum of series factorial(2*n)/4^n

The solution

You have entered [src]

  oo        
____        
\   `       
 \    (2*n)!
  \   ------
  /      n  
 /      4   
/___,       
n = 1

$$\sum_{n=1}^{\infty} \frac{\left(2 n\right)!}{4^{n}}$$

Sum(factorial(2*n)/4^n, (n, 1, oo))

The rate of convergence of the power series

The answer [src]

  oo            
 ___            
 \  `           
  \    -n       
  /   4  *(2*n)!
 /__,           
n = 1

$$\sum_{n=1}^{\infty} 4^{- n} \left(2 n\right)!$$

Sum(4^(-n)*factorial(2*n), (n, 1, oo))

Numerical answer

The series diverges

The graph

Sum of series factorial(2*n)/4^n

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
```