Prove that (1 + 1/n)^n ﹤ 3 (ILIEKMATHPHYSICS)

In this video, we prove that for all positive integers n, (1 + 1/n)^n ﹤ 3. This is a useful component if you want to prove that the sequence (1 + 1/1)^1, (1 + 1/2)^2, (1 + 1/3)^3, ..., (1 + 1/n)^n, ... converges. In fact, this sequence converges to the value e = 2.718281828459.... This proof shows that this sequence is bounded above (by 3), but we did not show that the sequence is monotonically increasing. In fact, you can use a very similar argument (using Bernoulli's Inequality) to show that this sequence is monotonically increasing. By the Monotone Convergence Theorem, the sequence converges, and we may call the value it converges to by "e".

Notice that the method to prove (1 + 1/n)^n ﹤ 3 in this video can be extended to a stronger claim -- in the sense that you can also prove (1 + 1/n)^n ﹤ 2.75, or even (1 + 1/n)^n ﹤ 2.72.

Thanks and enjoy the video!

[pehliks]

I like this style of video!! Its nice that you speed up the 'uninteresting' parts, and you are good at explaining your arguments in a short and concise way

[salem]

The limit is e as n approaches infinity. Great proof

[SbFH]

Since limit of (1+1/x)^n, as n→∞, is e. And, 2 < e < 3. Since n is finite (n < ∞), intuitively it could be concluded that even for larger n’s, it is less than 3, so for smaller n's it ofcourse follows that.

[SigmaChuck]

With a little calculus you can get a simpler argument: It suffices to show that 3^(1/n) > 1/n + 1 for any n in Z+. But this follows from 1) 3^(1/x) is monotonically decreasing on (0, inf) since its derivative (-ln 3)(3^(1/x))/x^2 is negative on positive R. And 2) for n=1, 3^(1/1) > 1/1+1 = 2, which is obviously true. Thus 3^(1/n) > 1/n+ 1 for n>=1, => 3= (3(1/n))^n) > (1+1/n)^n since 1+1/n and 3 are both positive real numbers greater than one. QED

[RexxSchneider]

Simply do the binomial expansion of (1+x/n)^n = 1 + nx/n + n(n-1)(x/n)^2/2! + n(n-1)(n-2)(n-3)(x/n)^3/3! + ... + n!(x/n)^n/n!.
That is term-by-term less than 1 + x + x^2/2! + x^3/3! + ... x^n/n! which is less than e^x.
So setting x = 1, we find (1+1/n)^n < e^1 = e < 3.

[ralfbodemann]

Nice proof, indeed! I expected you to apply induction.

[smithfrederick]

thanks for the vids, in my final year of math undergrad and love just keeping my thunker fresh

[roderictaylor]

As I suspect you're aware, a similar argument shows (1+1/n)^n is an increasing sequence. Together with the result (1+1/n)^(n+1) is a decreasing sequence, the first sequence is bounded above by any element of the second sequence and vice versa, so both sequences have a limit, and since the ratio of the two sequences goes to 1, they have a common limit.

[renesperb]

I suggest a different proof: consider f[x]= ln[1+1/x)^x=x*ln(1+1/x] .It is not difficult to show f[x] is increasing in x for x≥ 1.
If we denote y = 1/x then we can use De L'Hopital's rule to see that lim f[y] for y -> 0 = 1 =lim f[x], x-> inf. Hence ln((1+1/x)^x )<1
for x>1 that means that for any n>1 (1+1/n)^n < e <3.

[bjornfeuerbacher]

You don't actually need the Bernoulli inequality for proving the lemma. You can also do it this way:

((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) = ((n²+2n+1)/(n²+2n))^(n+1) (n+1)/(n+2)
= (n²+2n+1)/(n²+2n) (n²+2n+1)/(n²+2n) (n²+2n+1)/(n²+2n) (n+1)/(n+2),
where the first factor appears (n+2) times in the product. Now we use that this factor is obviously > 1 and that for such a fraction (with positive numerator and denominator), the value of the fraction becomes greater if we decrease both the numerator and the denominator by 1. So we have
((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2)
> (n²+2n+1)/(n²+2n) (n²+2n)/(n²+2n-1) (n²+n+1)/(n²+n) (n+1)/(n+2),
and now in this product of all the (n+1) fractions, lots of the denominators and numerators will cancel, leaving us with
((n+1)/n)^(n+1) / ((n+2)/(n+1))^(n+2) > (n²+2n+1)/(n²+n) (n+1)/(n+2) = (n²+2n+1)(n²+2n) > 1.

[ADDiOUMAARIR]

Great proof my man, you are the goat ❤❤

[henrikholst]

Mathematics often puts up a lot of scaffolding which they then remove and just magically pull out og the hat in the proof. Short and sweet but it hides the chain of though for us. In a previous video he used the term "scratch work" and this is exactly the kind of scaffolding I'm talking about.

[Jim-besj]

Good one. Did anybody work out a proof by induction? I tried for a while and gave up and theorems about convergent monotonic sequences to show it.

[robinwang]

Can we do log base n on both sides then expand the log that remains using a Taylor series, and say something about the left side looking like a truncated Taylor series?

[phnml]

Why do these videos kinda give me a 2000s vibe? And why do I like it?

[conjurer]

why did you create a seperate case for numbers less then 5? is there some rule that didn't work for them?
edit: ah right they are more then three when you increase the power by 1 so your proof doesn't work for them

[sn]

Why is n=5? The statements seem to be chosen completely random: e.g. it's somehow ^(n+1) and not ^n in the lemma. I always feel discouraged to learn math after seeing such things...

[WillhelmLiebniz]

I think your lemma counts as the induction step and then you just need to show the base case of n = 1 giving us less than 3 and you can claim by induction hypothesis the theorem is true.

[johnvandenberg]

Are you sure this is the easiest way to prove the theorem? 😁

[alexandreaussems]

Loved it, did you choose 5 because it's the first number where (1 + 1/n)^{n + 1} is less then 3 ?