Question

Evaluate the following limit:
\[\lim_{n \rightarrow \infty}(\sum_{i=1}^{4,000}i^\frac{1}{n}-3,999)^n\]

Answer 1

im trying to mess with below limit bu the sum looks complicated to simplify..
\[\large e^x = \lim \limits_{n\to \infty} \left(1 + \dfrac{x}{n}\right)^n\]

Answer 2

\[\lim_{n \rightarrow \infty}(\sum_{i=1}^{4,000}i^\frac{1}{n}-3,999)^n\]

Notice we can write the above as:

\[\lim_{n \rightarrow \infty}((1^\frac{1}{n}-1)+(2^\frac{1}{n}-1)+(3^\frac{1}{n}-1)+ \cdots + (4000^\frac{1}{n}-1)+1)^n\]

this a^z-1 thing should look kind of familiar

\[\text{ Let } T(x)=a^x\]
\[\text{ So } T'(x)=a^x \ln(a) =\lim_{z \rightarrow x} \frac{a^z-a^x}{z-x} \]
\[\text{ Then } T'(0)=ln(a)=\lim_{z \rightarrow 0} \frac{a^z-1}{z}\]
Ok do you see that a^z -1
we can use that for our problem if we let z=1/n
so
\[\ln(a)=\lim_{n \rightarrow \infty } \frac{a^\frac{1}{n}-1}{\frac{1}{n}} \]

For large n there exist small epsilon such that
\[\ln(a)=\frac{a^\frac{1}{n}-1}{\frac{1}{n}} + \epsilon \]

This still holds if we multiply both sides by 1/n
\[\frac{1}{n} \ln(a)=a^\frac{1}{n}-1+ \frac{1}{n} \epsilon \]

since 1/n*epsilon approaches 0 for large n we can use the approximation:
\[\frac{1}{n} \ln(a) \approx a^\frac{1}{n}-1\]
So let's look at the thing we rewrote earlier :
\[\lim_{n \rightarrow \infty}((1^\frac{1}{n}-1)+(2^\frac{1}{n}-1)+(3^\frac{1}{n}-1)+ \cdots + (4000^\frac{1}{n}-1)+1)^n\]

\[\lim_{n \rightarrow \infty}(\frac{1}{n} \ln(1)+\frac{1}{n} \ln(2)+\frac{1}{n} \ln(3)+ \cdots +\frac{1}{n} \ln(4000)+1)^n \]
\[\lim_{n \rightarrow \infty}(\frac{1}{n} \ln(4000!)+1)^n \]

\[\text{ Let } y=(\frac{\ln(4000!)}{n}+1)^n \]
\[\lim_{n \rightarrow \infty} \ln(y)=\lim_{n \rightarrow \infty} \frac{\ln(1+\frac{\ln(4000!)}{n})}{\frac{1}{n} }=\lim_{n \rightarrow \infty} \frac{\frac{\frac{-\ln(4000!)}{n^2}}{1+\frac{\ln(4000!)}{n}}}{-\frac{1}{n^2}} \\ =\lim_{n \rightarrow \infty}\frac{\frac{-\ln(4000!)}{n^2+\ln(4000!)n}}{\frac{-1}{n^2}} \\ =\lim_{n \rightarrow \infty}\frac{\ln(4000!)n^2}{n^2+\ln(4000!)n}=\ln(4000!)\]
Therefore \[\lim_{n \rightarrow \infty}(\sum_{i=1}^{4,000}i^\frac{1}{n}-3,999)^n\]
=
\[e^{\ln(4000!)}=4000!\]

Answer 3

yeah that is the exact formula I used

Answer 4

I used an approximation.

Answer 5

If anyone has any other way of looking at this let me know.

Answer 6

I usually play with small numbers first.

Answer 7

you know that have this form

Answer 8

beautiful !! i wouldn't have gotten this clever idea : \(\large \frac{1}{n} \ln(a) \approx a^\frac{1}{n}-1\)

i think the number 3999 exists exactly for this purpose xD

Answer 9

I remember a very similar problem to this
I had a lot of trouble with it
I think it was on the Putnam exam 7 years ago (maybe more years ago)
It is weird how some things you don't get until way later

Answer 10

\[\lim_{n \rightarrow \infty}(\sum_{i=1}^{k}i^\frac{1}{n}-(k-1))^n=k!\]

Answer 11

@ganeshie8 why did you think that one formula would be involved? just curious. what was your thinking?

Answer 12

was it simply because of the ( )^n that you thought it

Answer 13

Oh, i have worked a lot on my notes - since the n is in the exponent, outside the sum, I thought of simplifying the sum first before taking the limit.... i got stuck at that part... once the sum simplifies, rest of the problem is trivial

Answer 14

lol exactly ^

Answer 15

i think you did a lot better than me seeing this problem for the first time
when i seen it i was so completely lost at first

Answer 16

follow *

Answer 17

WOW! This is ridiculous man!

Answer 18

i cant think of anything better that was amazing approach !