A sequence { a n } is defined by a 1 = k and a n + 1 = a n − ( n + 1 ) ! 1 for all positive integers such that k is a constant number.
Also, m = 1 ∑ ∞ a m = K ( K is a constant number)
Find the sum of k and K .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
a n = k − ∑ i = 2 n i ! 1
As can be proven by induction. Therefore
lim n → ∞ a n = k + 2 − e Since e = ∑ i = 0 ∞ i ! 1
The partial sum is by induction
S n = n ( k − ∑ i = 2 n i ! 1 ) + ∑ i = 2 n i ! i − 1 = n a n + ∑ i = 2 n i ! i − 1 .
For the sum to comverge first term of right end side must be zero. Hence k = e − 2
In this way:
lim n → ∞ n a n = ∑ i = n ∞ i ! n = 0
In order to evaluate the 2nd term use the identity
e x = ∑ i = 0 ∞ i ! x i
Divide the identity by x and take the first derivative:
x e x − x 2 e x = − x 2 1 + ∑ i = 2 ∞ i ! ( i − 1 ) x i − 2
Using x = 1 in the above expression yelds
0 = − 1 + ∑ i = 2 ∞ i ! ( i − 1 )
1 = ∑ i = 2 ∞ i ! ( i − 1 )
Therefore
K = lim n → ∞ S n = 1
K + k = e − 1 = 1 . 7 1 8
Problem Loading...
Note Loading...
Set Loading...
claim : a n = k − s = 2 ∑ n s ! 1 ∀ n ≥ 2
proof(by induction) : the base case of n = 2 holds true. a n + 1 = a n − ( n + 1 ) ! 1 = k − ( s = 2 ∑ n s ! 1 ) − ( n + 1 ) ! 1 = k − s = 2 ∑ n + 1 s ! 1 and hence by induction this is proven.
for the series to converge, we need n → ∞ lim a n = 0 → k = n → ∞ lim s = 2 ∑ n s ! 1 = e − 2 → a n = e − 2 − s = 2 ∑ n s ! 1 we can make the sum nicer by using the taylor series of e x evaluated at x = 1 . so m = 1 ∑ ∞ ( e − 2 − s = 2 ∑ m s ! 1 ) = m = 1 ∑ ∞ s = m + 1 ∑ ∞ s ! 1 = s = 2 ∑ ∞ m = 1 ∑ s − 1 s ! 1 = s = 2 ∑ ∞ s ! s − 1 = s = 2 ∑ ∞ ( ( s − 1 ) ! 1 − s ! 1 ) = 1