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e: lim = \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right)

\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right) since:

(1) Since, it can be shown that both  \lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n and \lim\limits_{n \to \infty}\left(\frac{1}{n!}\right), the only remaining point is to show that:

\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n =\lim\limits_{n \to \infty}1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \dots + + \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right)  = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}

 

Is \lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n greater than, less than, or equal to \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right)

\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n converges

\lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n converges since:

(1)  From the binomial theorem:  

\left(1+\frac{1}{n}\right)^n = \sum\limits_{k=0}^{n}{n \choose k}\left(\frac{1}{n}\right)^{k} = =1 + \frac{n!}{(n-1)!}\left(\frac{1}{n}\right) + \frac{n!}{(n-2)!2!}\left(\frac{1}{n}\right)^2 + \frac{n!}{(n-3)!3!}\left(\frac{1}{n}\right)^3 + \dots + \frac{n!}{n!}\left(\frac{1}{n}\right)^n

=1 + 1 + \frac{n-1}{2!}\left(\frac{1}{n}\right) +  \frac{(n-1)(n-2)}{3!}\left(\frac{1}{n}\right)^2 + \dots + \frac{(n-1)(n-2)\dots(2)}{n!}\left(\frac{1}{n}\right)^{n-1}

=1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \dots + + \frac{1}{n!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{n-1}{n}\right)  \le \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}

(2) From this result, we know that \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\dfrac{1}{i!}\right) converges.


Does \lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n converge or diverge?

Binomial Theorem: \sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n

\sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n since:

(1)  For n=0, \sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = {0 \choose 0}x^{0}y^{0} = 1 = (x+y)^0

(2)   Assume that \sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k} = (x+y)^n 

(3)  (x+y)(x+y)^n = = (x+y)(x^n + {n \choose 1}x^{n-1}y + {n \choose 2}x^{n-2}y^2 + \dots + {n \choose k}x^{n-k}y^k + + \dots {n \choose {n-1}}xy^{n-1} + y^n) 

= x^{n+1} + \left[1 + {n \choose 1}\right]x^ny + \left[{n \choose 1} + {n \choose 2}\right]x^{n-1}y^2 + \dots \left[{n \choose k-1} + {n \choose k}\right]x^{n-k+1}y^k + \dots + \left[{n \choose {n-1}} + 1\right]xy^n + y^{n+1}

(4) Using Pascal's Rule {n \choose {r-1}} + {n \choose r} = {{n+1} \choose r} and 1 = {n \choose 0} = {n \choose n} so that:

x^{n+1} + \left[1 + {n \choose 1}\right]x^ny + \left[{n \choose 1} + {n \choose 2}\right]x^{n-1}y^2 + \dots \left[{n \choose k-1} + {n \choose k}\right]x^{n-k+1}y^k + \dots + \left[{n \choose {n-1}} + 1\right]xy^n + y^{n+1} =

{{n+1} \choose 0}x^{n+1} + {{n+1} \choose 1}x^ny + {{n+1} \choose 2}x^{n-1}y^2 + \dots {{n+1} \choose k}x^{n-k+1}y^k + \dots + {{n+1} \choose {n-1}}xy^n + {{n+1}\choose {n+1}}y^{n+1} = \sum\limits_{k=0}^{n+1}{{n+1} \choose k}x^ky^{n-k}  

\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) converges

\sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) converges since:

(1) \sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dots + \dfrac{1}{n!} =

= 1 + 1 + \frac{1}{1\times{2}} + \frac{1}{1\times{2}\times{3}} + \dots + \frac{1}{1\times{2}\times{3}\times\dots{n}} <

< 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}}    

(2)  Let u = 1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots + \dfrac{1}{2^{n-1}} 

(3)  u - \dfrac{1}{2}u = \dfrac{1}{2}u = 1 - \dfrac{1}{2^n} so that u = 2 - \dfrac{1}{2^{n-1}} < 2

(4)  So it follows that \sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) < 1+2=3

Does \sum\limits_{n=0}^\infty\left(\dfrac{1}{n!}\right) converge or diverge?

Pascal's Identity: {n \choose {r-1}} + {n \choose r} = {{n+1} \choose r}?

Here's the proof:

(1)  The definition of the binomial coefficient is:
{n \choose r} = \frac{n(n-1)\times\dots\times(n-r+1)}{r!}

(2)  This analysis will assume that n,r are integers.

(3)  If r > n+1, it follows that Pascal's Identity is trivially true since 0 = {n \choose {r-1}} = {n \choose r} = {{n+1} \choose r} so we can assume that r \le n+1

(4) {n \choose {r-1}} + {n \choose r} = \dfrac{n!}{(r-1)!(n-r+1)!} + \dfrac{n!}{r!(n-r)!} =
= n!\left(\frac{r}{r!(n-r+1)!}+\frac{n-r+1}{r!(n-r+1)!}\right) = \frac{n!(n+1)}{r!(n-r+1)!} = {{n+1} \choose r}

What is {n \choose {r-1}} + {n \choose r}?

Is \left(\sum\limits_{i=1}^n \left(\dfrac{1}{i}\right) - \log n\right) convergent or divergent?

e: \lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right)

\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = \lim\limits_{n \to \infty}\sum\limits_{i=0}^n\left(\frac{1}{i!}\right) since: (...