A special sequence

Calculus Level 5

A sequence { a n } \left\{ { a }_{ n } \right\} is defined by a 1 = k { a }_{ 1 }=k and a n + 1 = a n 1 ( n + 1 ) ! { a }_{ n+1 }={ a }_{ n }-\dfrac { 1 }{ \left( n+1 \right) ! } for all positive integers such that k k is a constant number.

Also, m = 1 a m = K \displaystyle\sum _{ m=1 }^{ \infty }{ { a }_{ m } } =K ( K K is a constant number)

Find the sum of k k and K K .


The answer is 1.71828.

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2 solutions

Aareyan Manzoor
Jul 28, 2019

claim : a n = k s = 2 n 1 s ! n 2 \text{claim}:a_n= k-\sum_{s=2}^n \dfrac{1}{s!} \forall n\geq 2

proof(by induction) : the base case of n = 2 n=2 holds true. a n + 1 = a n 1 ( n + 1 ) ! = k ( s = 2 n 1 s ! ) 1 ( n + 1 ) ! = k s = 2 n + 1 1 s ! a_{n+1} = a_n-\dfrac{1}{(n+1)!} =k-\left(\sum_{s=2}^n \dfrac{1}{s!}\right)-\dfrac{1}{(n+1)!}= k-\sum_{s=2}^{n+1} \dfrac{1}{s!} and hence by induction this is proven.

for the series to converge, we need lim n a n = 0 k = lim n s = 2 n 1 s ! = e 2 a n = e 2 s = 2 n 1 s ! \lim_{n\to \infty} a_n = 0 \to k= \lim_{n\to \infty} \sum_{s=2}^n \dfrac{1}{s!} = \boxed{ e-2}\to a_n =e-2-\sum_{s=2}^n \dfrac{1}{s!} we can make the sum nicer by using the taylor series of e x e^x evaluated at x = 1 x=1 . so m = 1 ( e 2 s = 2 m 1 s ! ) = m = 1 s = m + 1 1 s ! = s = 2 m = 1 s 1 1 s ! = s = 2 s 1 s ! = s = 2 ( 1 ( s 1 ) ! 1 s ! ) = 1 \sum_{m=1}^\infty \left(e-2-\sum_{s=2}^m \dfrac{1}{s!}\right) = \sum_{m=1}^\infty \sum_{s=m+1}^\infty \dfrac{1}{s!}= \sum_{s=2}^\infty \sum_{m=1}^{s-1} \dfrac{1}{s!}= \sum_{s=2}^\infty\dfrac{s-1}{s!} = \sum_{s=2}^\infty\left(\dfrac{1}{(s-1)!}-\dfrac{1}{s!}\right) = \boxed{1}

Matteo Di Prinzio
Jul 29, 2019

a n = k i = 2 n 1 i ! \ a_{n}=k-\sum_{i=2}^n \frac{1}{i!}

As can be proven by induction. Therefore

lim n a n = k + 2 e \lim_{n \rightarrow \infty} \ a_{n}=k+2-e Since e = i = 0 1 i ! e=\sum_{i=0}^{\infty} \frac{1}{i!}

The partial sum is by induction

S n = n ( k i = 2 n 1 i ! ) + i = 2 n i 1 i ! = n a n + i = 2 n i 1 i ! \ S_{n}=n(k-\sum_{i=2}^n \frac{1}{i!})+\sum_{i=2}^n \frac{i-1}{i!}=na_{n}+\sum_{i=2}^n \frac{i-1}{i!} .

For the sum to comverge first term of right end side must be zero. Hence k = e 2 \ k=e-2

In this way:

lim n n a n = i = n n i ! = 0 \lim_{n \rightarrow \infty} na_{n}=\sum_{i=n}^{\infty} \frac{n}{i!}=0

In order to evaluate the 2nd term use the identity

e x = i = 0 x i i ! e^x=\sum_{i=0}^{\infty} \frac{x^i}{i!}

Divide the identity by x \ x and take the first derivative:

e x x e x x 2 = 1 x 2 + i = 2 ( i 1 ) x i 2 i ! \frac{e^x}{x}-\frac{e^x}{x^2}=-\frac{1}{x^2}+\sum_{i=2}^{\infty} \frac{(i-1)x^{i-2}}{i!}

Using x = 1 \ x=1 in the above expression yelds

0 = 1 + i = 2 ( i 1 ) i ! 0=-1+\sum_{i=2}^{\infty} \frac{(i-1)}{i!}

1 = i = 2 ( i 1 ) i ! 1=\sum_{i=2}^{\infty} \frac{(i-1)}{i!}

Therefore

K = lim n S n = 1 K= \lim_{n \rightarrow \infty} \ S_{n}=1

K + k = e 1 = 1.718 K+k =e-1=\boxed{1.718}

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