# 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,

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

**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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