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

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

(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:

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

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

Simplify $\sum\limits_{k=0}^{n}{n \choose k}x^ky^{n-k}$

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

Proof that $\sum\limits_{i=1}^n \left(\dfrac{1}{i}\right)$ is divergent

 (1)  Assume that $\sum\limits_{i=1}^n \left(\dfrac{1}{i}\right)$ is convergent so that there exists $H$ such that:

$$H = \sum\limits_{i=1}^n \left(\dfrac{1}{i}\right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$$

(2)  $H \ge 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{8} + \dots =$

$$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots = \frac{1}{2} + H$$

(3)  Since we have reached a contradiction, we can reject the assumption and conclude that the sequence is divergent.

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

What is the value of $\lim\limits_{n \to \infty}\left(1+\dfrac{x}{n}\right)^n$ (with answer)

$\lim\limits_{n \to \infty}\left(1+\dfrac{x}{n}\right)^n = e^x$ since:

(1)  For $x = 0$, $\lim\limits_{n \to \infty}\left(1+\dfrac{0}{n}\right)^n =  1 = e^0$

(2)  So we can assume that $x \ne 0$ so that:

$$\lim\limits_{n \to \infty}\left(1+\dfrac{x}{n}\right)^n = \lim\limits_{n \to \infty}e^{n\ln\left(1+\dfrac{x}{n}\right)} = \lim\limits_{n \to \infty}e^{x\ln\left(\frac{1+x/n}{x/n}\right)}$$

(3)  Since $\lim\limits_{n \to \infty}\dfrac{1+x/n}{x/n} = \lim\limits_{h \to 0}\dfrac{1+h}{h}$:

$$\lim\limits_{n \to \infty}e^{x\ln\left(\frac{1+x/n}{x/n}\right)} = \lim\limits_{h \to 0}e^{x\ln\left(\frac{1+h}{h}\right)} = e^{x\lim\limits_{h \to 0}\left(\ln\left[\frac{1+h}{h}\right]\right)}$$

(4)  Since $\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}=1$ (see here if needed):

$$e^{x\lim\limits_{h \to 0}\left(\ln\left[\frac{1+h}{h}\right]\right)} = e^x$$

What is the value of $\lim\limits_{n \to \infty}\left(1+\dfrac{x}{n}\right)^n$

What is the value of $\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}$ (with answer)

$\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}=1$ since:

(1) $\ln(a) - \ln(b) = \ln\left(\dfrac{a}{b}\right)$ since:

• Let $y = \ln(a) - \ln(b)$
• $e^y = e^{\ln(a) - \ln(b)} = \dfrac{e^{\ln(a)}}{e^{\ln(b)}} = \dfrac{a}{b}$
• $y = \ln\left(\dfrac{a}{b}\right)$

(2) $\lim\limits_{h \to 0}\dfrac{\ln(x+h) - \ln(x)}{h} = \lim\limits_{h \to 0}\dfrac{\ln\left(1 + \frac{h}{x}\right) }{h}$

(3)  Let $t = \dfrac{h}{x}$ so that $h = xt$ and:

$$\lim\limits_{h \to 0}\frac{\ln\left(1 + \frac{h}{x}\right) }{h} = \lim\limits_{t \to 0}\frac{\ln\left(1 + t\right) }{xt} = \lim\limits_{t \to 0}\left(\frac{1}{xt}\right)\ln(1+t)$$

(4)  $n\ln(a) = \ln\left(a^n\right)$ since:

• Let $y = n\ln(a)$
• $e^y = e^{n\ln(a)} = \left(e^{\ln(a)}\right)^n = a^n$
• $y = \ln\left(a^n\right)$

(5)  $\lim\limits_{t \to 0}\left(\frac{1}{xt}\right)\ln(1+t) = \lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{xt}}\right) = \lim\limits_{t \to 0}\ln\left(\left[\left(1+t\right)^{\frac{1}{t}}\right]^{\frac{1}{x}}\right) = \left(\dfrac{1}{x}\right)\lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{t}}\right)$

(6)  By definition, $e = \lim\limits_{n \to \infty}\left(1 + \dfrac{1}{n}\right)^n$ so that if $u=\dfrac{1}{t}$, then:

$$ \left(\frac{1}{x}\right)\lim\limits_{t \to 0}\ln\left(\left[1+t\right]^{\frac{1}{t}}\right) =\left(\frac{1}{x}\right)\lim\limits_{u \to \infty}\ln\left(\left[1+\frac{1}{u}\right]^{u}\right) =\left(\frac{1}{x}\right)\ln(e) = \frac{1}{x}$$

(7)  Since $\lim\limits_{h \to 0}\dfrac{\ln(x+h) - \ln(x)}{h} = \dfrac{1}{x}$, it follows that:

$$\lim\limits_{h \to 0}\frac{\ln(1+h)}{h} = \lim\limits_{h \to 0}\frac{\ln(1+h) - \ln(1)}{h} = \frac{1}{1}=1$$

What is the value of $\lim\limits_{h \to 0}\dfrac{\text{ln}(1+h)}{h}$