Studying Mathematics Thread

[GrandizerII]

The limit (as n approaches infinity) of (1+2+3+...+n)/(n^2) is actually 1/2 ... not 0!

I was confused when I first saw this because just looking at it, it seemed that you just had to increase the value of n to infinity (in which case the limit would be 0), but what I initially failed to keep in mind is that not only is n increasing in value as it approaches infinity but the number of terms increases in the numerator as well. That's why the denominator never goes high enough to demolish the fraction to 0/infinity.

As to how to arrive at the correct answer:

the numerator (1+2+3+...+n) = n(n+1)/2
I will demonstrate this is true another time ...

This means the whole expression up above (after the limit) is n(n+1)/(2n^2) = (n+1)/(2n) = n/(2n)+1/(2n) = 1/2 + 1/(2n)

Therefore, the limit (as n approaches infinity) of that result is 1/2.

OK, now try

(1^2 + 2^2 + 3^2 +...+n^2)/n^3

and

(1^3 + 2^3 + 3^3 +...+n^3 )/n^4.

Any guesses for a generalization?

First series:

This is the theorem for the numerator series:

1^2 + 2^2 + 3^2 +...+n^2 = n(n+1)(2n+1)/6

Including the denominator, this means the first series up above in the quote is equal to:

n(n+1)(2n+1)/(6n^3) = (n+1)(2n+3)/(6n^2) = (2n^2+5n+3)/(6n^2) = 1/3 + 5/(6n) + 1/(2n^2)

The limit (as n approaches infinity) of n(n+1)(2n+1)/(6n^3) is therefore:

1/3

Second series:

1^3 + 2^3 + 3^3 +...+n^3 = (n^2(n+1)^2)/4

Including the denominator, this means the second series up above in the quote is equal to:

(n^2(n+1)^2)/(4n^4) = (n^2+2n+1)/(4n^2) = 1/4 + 1/(2n) + 1/(4n^2)

The limit (as n approaches infinity) of (n^2(n+1)^2)/(4n^4) is therefore:

1/4