# 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?

$\begin{eqnarray} 1 + 2 + 3 + \cdots + n &=& \dfrac12n(n+1) \\ 1^3 + 2^3 + 3^3 + \cdots + n^3 &=& \left( \dfrac12n(n+1) \right)^2 \\ \end{eqnarray}$

Step 2: There is a relationship between the numbers $$1 + 2 + 3 + \cdots + n$$ and $$1^3 + 2^3 + 3^3 + \cdots + n^3$$,

$(1 + 2 + 3 + \cdots + n )^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3$

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

$\lim_{n\to\infty} (1 + 2 + 3 + \cdots + n )^2 = \lim_{n\to\infty} (1^3 + 2^3 + 3^3 + \cdots + n^3)$

Which is equivalent to

$(1+2+3 + \cdots )^2 = 1^3 +2^3 + 3^3 + \cdots$

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

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

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