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By the way, do you know the RUclips channel of Dr Peyam? He has a whole playlist on the Gaussian integral, with 12 different methods. But yours was not included there, so apparently it's really a new one?

Yes, I'm a fan of Dr Peyam! This approach is adapted from/related to something I remember seeing a few years ago - I can't seem to find a reference though for where I found it. I'll see if I can find it again.

The 3n must be inside grouping symbols: lim n*[1 + 1/(3n) - 1] = lim n*1/(3n) = 1/3

@@robertveith6383 it's not 1/3*n 3n is grouped automatically

This is how I did it

That's how I did it!

I used the binomial expansion too but I didn't factor out the n^3 first which made it a little more interesting. I think I know what you mean - I wouldn't have thought that putting a sum of 1/3 and 2/3 power in the denominator would simplify things.

@@dugong369 the (1+epsilon)^n expansion is very well known at least in physics context.

@@ikarienator yes my way was less rigorous as well as a little more complicated. To be even sloppier, assume the limit is x. Then n^3+n^2 = (n+x)^3 in the limit. n^3+n^2 = n^3+3xn^2+3nx^2+x^3. In order for the 2 highest order terms to have matching coefficients, x=1/3.

It doesn't fail. Use a series expansion on n((1+1/n)^(1/3) - 1) and then cancel the 1's and muiltiply through by the n. You get (1/3 - terms in 1/n ...). The limit as n → ∞ is obviously 1/3.

@@RexxSchneider of course - a long day 🙂

Saar ! Do not redeem !

​@@Regimeducamp I make bangers, not anthems, leave that to the artful dodgers.

@@brianstine2006 It's interesting that adding the n/3 + 1/27 contribution doesn't affect the limit, so while the n^3 term doesn't dominate at infinity, it seems like n^3 + n^2 does!

The explicit n^3 doesn't dominate at infinity because, being under the cube root sign, it only grows like n and is cancelled by the -n. Consider, e.g., pulling a factor of n out of the entire expression to get n((1+1/n)^{1/3} - 1), and looking at the taylor expansion of the binomial under the cube root, where the first term of the taylor expansion gets cancelled by the -1 at the end. A lot of confusing limits like this involve the first term(s) of some taylor expansions cancelling each other so only higher-order terms survive.

I confused shouldnt the highest order term in the root match the order of the root?

Try checking that for m = 1 lim_{n to infinity} ( (n + n^2)^{1/1} - n ) = 1/1 => lim_{n to infinity} ( n + n^2 - n ) = 1 => lim_{n to infinity} ( n^2 ) = 1 ???

Sorry, typo, I've corrected it now.

...or using Taylor expansion, which makes the result immediate and trivial

That's indeed what chatgpt came up with for when I tried the general case the other guy above mentioned

For the last step we can also note that, by definition, lim ((1+x)^(1/3)-1)/x, x->0 is the derivative of t^(1/3) at t = 1.

@@vladimir10 Use a binomial expansion and you don't need any calculus. (1 + 1/n)^m = (1 + m(1/n) + m(m-1)/2!(1/n)^2 ... ) Set m=1/3 for this case.

Interesting! What values did you use for b and c?

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996 b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497 c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125 We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

@@DrBarker I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later.

@@DrBarker I found a whole set of solutions now with quite a few different values of a, b, and c. a can range from 1 to 996. For each value of a, the terms b and c can be algorithmically determined. a cannot be 997 or higher because 997 is prime, and once two of them have been removed from the expression, the remaining two cannot be represented by any combination of b and c. I started out the same way, transforming the expression into the product of n raised to the power of (1001-n) where n ranges from 2 to 1000, giving us our 999 terms. If a=996, the first 995 terms all get their exponents decreased by 2, and the a! squared term is accounted for. Now we just have to sort all our current terms into b and c. I reversed the order and counted down for my approach. 1000 contributes a multiplier of 10 to b. Our two 999's each become 3x333 and get sorted out later. Our three 998's go into b. Our four 997's go into c. Our remaining three 996's go into b. Our remaining four 995's go into c. We have five 994's left; we can either put three of them into b or four of them into c. I chose the latter for the one solution I fully worked out. That leaves one 994 that we turn into 2x497 and sort out later. All the remaining terms have exponents of 6 or greater. It is fairly quick and easy to see that any number greater than or equal to 6 can be represented as the sum of integer multiples of 3 and 4. 6=2x3+0x4, 7=1x3+1x4, 8=0x3+2x4, etc... When we get to numbers greater than integer multiples of 12, each block of 12 can get put into b four times or c three times, giving us our large variety of solutions. For the solution I worked out all the way, blocks of 12 always get put into c three times. At this point, our next term is 993 raised to the 6th power. It goes into b twice. Our seven 992's contribute once to b and once to c. Our eight 991's go into c twice. If we extend this pattern, each term contributes its n value to b, with its exponent cycling through a repeating sequence of (3,2,1,0,3,2,1,0...). Each term also contributes its n value to c, with its exponent cycling through a stair stepping sequence of (0,1,2,3,1,2,3,4,2,3,4,5,3,4,5,6...) We can apply this all the way down, making adjustments when we get to n=2, n=3, n=333, and n=497 in order to account for the modified exponents from earlier. This gives us the following: a=996, b=10x998x996xP1xP2, where P1 is the product of n^((n+1)mod4) for all n from 4 to 993 except 333 and 497, and P2=2^2x3^2x497, c=997x995x994xP3xP4, where P3 is the product of n^((994-n)mod4+the floor function of ((990-n)/4)) for all n from 4 to 993 except 333 and 497, and P4=2^249x3^249x333^167x497^125. We can repeat this process for any value of a from 1 to 996. We can also transform the expression into its prime factorization and then sort all the terms in this way in order to obtain the most complete set of solutions.

Thank you! That basically means an equation which we might solve to find a function, like how f(xy) = f(x) + f(y) has a solution f(x) = log(x), rather than the usual solving an equation to find the value of a variable like 2x + 1 = 5.

Bro I came here thinking the same thing 🤣.