you have sum[1,∞] 2n/n!
= 2 sum[1,∞] 1/(n-1)!
= 2 sum[0,∞] 1/n!
recall that e = 1 + 1/1! + 1/2! + 1/3!
so, your sum is 2e
Find the sum to infinity of the series
1 + 4/3! + 6/4! + 8/5!
2 answers
The series
1 + 4/3! + 6/4! + 8/5!
is
Σ2k/(k+1)! for k=1,∞
=2Σk/(k+1)! k=1,∞
=2Σ(k-1)/k! k=2,∞
=2Σk/k!-1/k! k=2,∞
=2Σ1/(k-1)! k=2,∞
- 2Σ1/k! k=2,∞
=2Σ1/k! k=1,∞
- 2(Σ1/k!-2) k=0,∞
=2(Σ1/k!-1) k=0,∞
- 2(e-2)
=2(e-1)-2(e-2)
=2
Noting that
e=Σ1/k! k=0,∞
1 + 4/3! + 6/4! + 8/5!
is
Σ2k/(k+1)! for k=1,∞
=2Σk/(k+1)! k=1,∞
=2Σ(k-1)/k! k=2,∞
=2Σk/k!-1/k! k=2,∞
=2Σ1/(k-1)! k=2,∞
- 2Σ1/k! k=2,∞
=2Σ1/k! k=1,∞
- 2(Σ1/k!-2) k=0,∞
=2(Σ1/k!-1) k=0,∞
- 2(e-2)
=2(e-1)-2(e-2)
=2
Noting that
e=Σ1/k! k=0,∞