What is the value of e when substituting n=infinity into the limit (1 + 1/n)^n?

[tahayassen]

I understand your post Steely Dan, but I was just wondering why I'm not getting the approximate value of 2 when I tried it myself:

http://i2.lulzimg.com/7009a29b9a.png

Tex version:

Ignoring\quad coefficients...\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }(1^{ n }{ ({ n }^{ -1 }) }^{ 0 }+{ 1 }^{ n-1 }{ ({ n }^{ -1 }) }^{ 1 }+{ 1 }^{ n-2 }{ ({ n }^{ -1 }) }^{ 2 }+\quad ...\quad +{ 1 }^{ 1 }{ ({ n }^{ -1 }) }^{ n-1 }+{ 1 }^{ 0 }{ ({ n }^{ -1 }) }^{ n })\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }(1+{ n }^{ -1 }+{ n }^{ -2 }+\quad ...\quad +{ n }^{ -n+1 }+{ n }^{ -n })\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }\left[ { n }^{ -n }({ n }^{ n }+{ n }^{ n-1 }+{ n }^{ n-2 }+\quad ...\quad +{ n }^{ 1 }+{ n }^{ 0 }) \right] \\ Ignoring\quad everything\quad except\quad the\quad first\quad two\quad terms...\\ { (1+{ n }^{ -1 }) }^{ n }={ \underset { n\rightarrow \infty }{ lim } }\left[ { n }^{ -n }({ n }^{ n }+{ n }^{ n-1 }) \right] \\ { (1+{ n }^{ -1 }) }^{ n }={ \infty }^{ -n }(\infty ^{ n }+{ \infty }^{ n-1 })

Results:

http://i2.lulzimg.com/912d8aca39.png

Am I doing something wrong? Do the coefficients matter? How did you get the coefficients of (n-1)?

 

[Steely Dan]

Am I doing something wrong? Do the coefficients matter? How did you get the coefficients of (n-1)?

The coefficients absolutely matter. In the way you wrote it, the second term is not (1)n^{-1} but (n)n^{-1}. If you're not seeing it, you have to actually work out the individual terms in the expansion.

 

[Mark44]

Regarding post #31, you should never substitute ∞ into an arithmetic expression.

In that post you have "∞^-n(∞ⁿ + ∞^-n)"

 

[tahayassen]

s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)

How did you get the coefficients: (n-1) for all the terms in the middle?

Wouldn't pascal's coefficients be something like:

http://i2.lulzimg.com/4084d10b33.png

 

[Steely Dan]

s_n = n^{-n} (1 + (n-1)n + ... + (n-1)n^{n-1} + n^n)

How did you get the coefficients: (n-1) for all the terms in the middle?

Wouldn't pascal's coefficients be something like:

http://i2.lulzimg.com/4084d10b33.png

That's correct. Obviously I didn't work out all of the coefficients, just the last three:

\left(\begin{array}{c} n \\ n \end{array}\right) = \frac{n!}{n!\ 0!} = 1
\left(\begin{array}{c} n \\ n-1 \end{array}\right) = \frac{n!}{(n-1)!\ 1!} = n
\left(\begin{array}{c} n \\ n-2 \end{array}\right) = \frac{n!}{(n-2)!\ 2!} = \frac{1}{2}(n)(n-1)

The coefficients are symmetric, so this would also be the first three.

 

[tahayassen]

Thanks! I understand now.

 

[tahayassen]

Actually, one last thing, if I look at your third term:

\frac { 1 }{ 2 } (n^{ 2 }-3n+2)n^{ n-2 }

But shouldn't it be:

\frac { 1 }{ 2 } (n)(n-1)n^{ n-2 }\\ =\frac { 1 }{ 2 } ({ n }^{ 2 }-n)n^{ n-2 }

 

[Steely Dan]

Yes, I tried doing the coefficients in my head when I wrote that and evidently made a mistake. Fortunately it didn't affect the leading term.