Find the sum to infinity of the series
1 + 4/3! + 6/4! + 8/5!
The book says the a answer is 2
sum[1,∞] 2n/(n+1)!
= 2 sum[1,∞] n/(n+1)!
= 2 sum[1,∞] (n+1-1)/(n+1)!
can you take it from there?
To find the sum to infinity of the given series, we need to determine if the series converges or diverges.
The sum to infinity of a series exists only if the series converges. If the terms of the series decrease to zero, then the series converges. If the terms do not approach zero, then the series diverges.
Let's analyze the terms of the given series to determine if it converges or diverges:
1. The first term is 1.
2. The second term is 4/3! = 4/6 = 2/3.
3. The third term is 6/4! = 6/24 = 1/4.
4. The fourth term is 8/5! = 8/120 = 1/15.
From the pattern observed above, we can see that the terms are decreasing as the series progresses. Since the terms are approaching zero, we can conclude that the series converges.
Now, let's find the sum to infinity of the series:
The nth term of the series is given by an = 2n / (n+1)!
To find the sum to infinity (S), we can use the formula for the sum of an infinite geometric series:
S = a1 / (1 - r)
where a1 is the first term and r is the common ratio.
In this series, a1 = 1 and the common ratio (r) can be found by taking the ratio of the n+1 term to the nth term:
r = (2(n+1) / ((n+2)!)) / (2n / (n+1)!)
= ((2n+2)(n+1)!) / ((n+2)!(2n))
= (2n+2) / (2n(n+2))
= (n+1) / (n+1)
= 1
Since r = 1, the common ratio is 1. Therefore, the sum to infinity (S) of the given series is:
S = a1 / (1 - r)
= 1 / (1 - 1)
= 1 / 0
The denominator 0 indicates that the sum to infinity of the series does not exist. Hence, the given series diverges.