Infinity is tricky to deal with

Calculus Level 3

I will attempt to prove that \( \infty = 1 \text{ or }\infty= -2\). In which of these steps did I first make a flaw in my logic?

Step 1: Using the algebraic identities, we have

\[ \begin{array} { l l l l l l } 1 &+ 2 &+ 3 &+ &\cdots &+ &n = \dfrac12 n(n+1) \\ 1^3 &+ 2^3 &+ 3^3 &+ &\cdots &+ &n^3 = \dfrac14 n^2(n+1)^2 \\ \end{array} \]

Step 2: Connecting these two equations to get

\[ (1+ 2 +3 + \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3. \]

Step 3: Since \(1 + 2 + 3 + \cdots = \infty\) and \(1^3 + 2^3 + 3^3 + \cdots = \infty\), then

\[ 1+ 2 +3 + \cdots + \infty = 1^3 + 2^3 + 3^3 + \cdots + \infty^3 . \]

Step 4: We can rewrite the equation in Step 3 as

\[ \dfrac12 \infty(\infty+1) = \left( \dfrac12 \infty(\infty+1) \right)^2 . \]

Step 5: Canceling the common factors gives

\[ 1 = \dfrac12 \infty (\infty + 1) . \]

Step 6: Solving this quadratic equation gives \( \infty =1 \) or \(\infty = -2\).

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