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Sum of series:
1/n^3
1/((3n-2)*(3n+1))
(x-1)^n/5^n
1/(4n^2-9)
Identical expressions
factorial(n)*n^(-p)*factorial(q- one)/factorial(q+n)
factorial(n) multiply by n to the power of ( minus p) multiply by factorial(q minus 1) divide by factorial(q plus n)
factorial(n) multiply by n to the power of ( minus p) multiply by factorial(q minus one) divide by factorial(q plus n)
factorial(n)*n(-p)*factorial(q-1)/factorial(q+n)
factorialn*n-p*factorialq-1/factorialq+n
factorial(n)n^(-p)factorial(q-1)/factorial(q+n)
factorial(n)n(-p)factorial(q-1)/factorial(q+n)
factorialnn-pfactorialq-1/factorialq+n
factorialnn^-pfactorialq-1/factorialq+n
factorial(n)*n^(-p)*factorial(q-1) divide by factorial(q+n)
Similar expressions
factorial(n)*n^(p)*factorial(q-1)/factorial(q+n)
factorial(n)*n^(-p)*factorial(q-1)/factorial(q-n)
factorial(n)*n^(-p)*factorial(q+1)/factorial(q+n)

Sum of series/ factorial(n)*n^(-p)*factorial(q-1)/factorial(q+n)

Sum of series factorial(n)n^(-p)factorial(q-1)/factorial(q+n)

The solution

You have entered [src]

  oo                 
____                 
\   `                
 \        -p         
  \   n!*n  *(q - 1)!
  /   ---------------
 /        (q + n)!   
/___,                
n = 1

$$\sum_{n=1}^{\infty} \frac{n^{- p} n! \left(q - 1\right)!}{\left(n + q\right)!}$$

Sum(((factorial(n)*n^(-p))*factorial(q - 1))/factorial(q + n), (n, 1, oo))

The answer [src]

  oo                  
____                  
\   `                 
 \     -p             
  \   n  *n!*(-1 + q)!
  /   ----------------
 /        (n + q)!    
/___,                 
n = 1

$$\sum_{n=1}^{\infty} \frac{n^{- p} n! \left(q - 1\right)!}{\left(n + q\right)!}$$

Sum(n^(-p)*factorial(n)*factorial(-1 + q)/factorial(n + q), (n, 1, oo))

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
```

Other calculators

How to use it?

Identical expressions

Similar expressions

Sum of series factorial(n)*n^(-p)*factorial(q-1)/factorial(q+n)

The solution

Examples of finding the sum of a series

Sum of series factorial(n)n^(-p)factorial(q-1)/factorial(q+n)