But it should be equal to -1/12

Calculus Level 3

Here's my proof that the sum of all the natural numbers is equal to 1.
In which of thse steps did I first make a flaw in my logic?

Step 1: Recall the two algebraic identities,

1+2+3++n=12n(n+1)13+23+33++n3=(12n(n+1))2 \begin{aligned} 1 + 2 + 3 + \cdots + n &=& \dfrac12n(n+1) \\ 1^3 + 2^3 + 3^3 + \cdots + n^3 &=& \left( \dfrac12n(n+1) \right)^2 \\ \end{aligned}

Step 2: There is a relationship between the numbers 1+2+3++n 1 + 2 + 3 + \cdots + n and 13+23+33++n31^3 + 2^3 + 3^3 + \cdots + n^3 ,

(1+2+3++n)2=13+23+33++n3 (1 + 2 + 3 + \cdots + n )^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3

Step 3: Take the limit to both sides of the equation,

limn(1+2+3++n)2=limn(13+23+33++n3) \lim_{n\to\infty} (1 + 2 + 3 + \cdots + n )^2 = \lim_{n\to\infty} (1^3 + 2^3 + 3^3 + \cdots + n^3)

Which is equivalent to

(1+2+3+)2=13+23+33+ (1+2+3 + \cdots )^2 = 1^3 +2^3 + 3^3 + \cdots

Step 4: Since 13+23+33+1^3 +2^3 + 3^3 + \cdots \rightarrow \infty and 1+2+3+1 +2 + 3 + \cdots \rightarrow \infty , then 13+23+33+=1+2+3+1^3 +2^3 + 3^3 + \cdots = 1 +2 + 3 + \cdots,

(1+2+3+)2=1+2+3+(1+2+3+)(1+2+3+)=1+2+3+(1+2+3+)(1+2+3+)=(1+2+3+)1+2+3+=1\begin{aligned} (1+2+3 + \cdots )^2 &=& 1 +2 + 3 + \cdots \\ (1+2+3 + \cdots )(1+2+3 + \cdots ) &=& 1 +2 + 3 + \cdots \\ \cancel{(1+2+3 + \cdots )}(1+2+3 + \cdots ) &=& \cancel{(1+2+3 + \cdots )} \\ 1+2+3 + \cdots &=& 1 \end{aligned}

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