infinite series

Question

sum the infinite series 1^2/0!+2^2/1!+3^2/2!+4^2/3!

Answer 1

Hi Tae,

 

The series sums to 5e. First, express the series as a summation:

 

S = Σ^∞_n=0 ( (n+1)²/n! )

 

Expand the numerator

 

S = Σ^∞_n=0 ( (n² + 2n + 1)/n! )

 

Write as seperate components:

 

S = Σ^∞_n=0 ( n²/n! + 2n/n! + 1/n! )

 

We know that Σ^∞_n=0 1/n! = e, so we can rewrite the above as

 

S = e + Σ^∞_n=0 ( n²/n! + 2n/n! )

 

Cancel out a few terms:

 

S = e + Σ^∞_n=1 ( n/(n-1)! + 2/(n-1)! )     (notice the index changes from n = 0 to n = 1)

 

Change the incex back to n = 0, thus increasing the n's in the summation by 1:

 

S = e + Σ^∞_n=0 ( (n+1)/n! + 2/n! )

 

Again, we know that Σ^∞_n=0 2/n! = 2 Σ^∞_n=0 1/n! = 2e, so we can rewrite the above as

 

S = e + 2e + Σ^∞_n=0 ( (n+1)/n! )

 

   = 3e + Σ^∞_n=0 ( n/n! + 1/n! )

 

Notice that we again have a 1/n! term in the summation, and since Σ^∞_n=0 1/n! = e, we have

 

S = 3e + e + Σ^∞_n=0 ( n/n! )

 

   = 4e + Σ^∞_n=1 ( 1/(n-1)! )     (Note again that the index changed from n = 0 to n = 1)

 

We change the index back to n = 0, thus increasing the n terms in the summation by 1:

 

S = 4e + Σ^∞_n=0 ( 1/n! )

 

   = 4e + e

 

   = 5e