Find the sum to infinity of the series
1 + 4/3! + 6/4! + 8/5!
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
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,∞
To find the sum to infinity of the series 1 + 4/3! + 6/4! + 8/5!, we need to determine if the series is convergent or divergent. If it is convergent, we can then find its sum.
First, let's rewrite the series using sigma notation:
∑ (n=1 to ∞) [2n / (n+1)!]
To determine the convergence of the series, let's apply the ratio test:
Given a series ∑ aₙ, if the limit as n approaches infinity of |aₙ₊₁ / aₙ| is less than 1, then the series converges. If it is greater than 1 or the limit does not exist, then the series diverges.
Let's apply the ratio test to our series:
lim (n→∞) [ | (2(n+1) / ((n+1)+1)! ) / (2n / (n+1)! ) | ]
Simplifying the expression:
lim (n→∞) [ (2(n+1) / ((n+1)+1)! ) * ((n+1)! / (2n) ) ]
The (n+1)! term cancels out:
lim (n→∞) [ (2(n+1) / (n+2)! ) * (n! / (2n) ) ]
Simplifying further:
lim (n→∞) [ (2(n+1) * n! ) / ( (n+2)! * (2n) ) ]
Now, let's simplify the factorials:
lim (n→∞) [ (2(n+1) * n! ) / ( (n+2)(n+1)! * (2n) ) ]
The (n+1)! term cancels out:
lim (n→∞) [ 2 / ( (n+2) * 2n ) ]
Simplifying further:
lim (n→∞) [ 1 / ( (n+2) * n ) ]
Taking the limit:
lim (n→∞) [ 0 ]
The limit is equal to 0.
Since the limit of |aₙ₊₁ / aₙ| is less than 1, the series is convergent.
To find the sum of the series, we can use the formula for the sum of a convergent series:
S = a / (1 - r)
where S is the sum, a is the first term, and r is the common ratio.
In our case, the first term a is 1, and the common ratio r is 1/2 (since 2n / (n+1)! = (2/n) * (1 / (1+1/n)) and as n approaches infinity, 1+1/n becomes 1).
Hence, the sum to infinity of the series 1 + 4/3! + 6/4! + 8/5! is:
S = 1 / (1 - 1/2)
= 1 / (1/2)
= 2
Therefore, the sum to infinity of the series is 2.