# Flawed alternatives

Calculus Level 5

It is known that the harmonic series $$\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\ldots$$ diverges. What about the alternating harmonic series $$\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\ldots$$?

Here, I will give a proof that the alternating harmonic series converges to 0!

Step 1: We rewrite the sum as follows.

$1-\frac{1}{2}+\frac{1}{3}-\ldots = \left( 1+\frac{1}{2}+\frac{1}{3}+\ldots \right) - 2 \left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots \right)$

Step 2: We express the two separated sums in sigma notation.

$\left( 1+\frac{1}{2}+\frac{1}{3}+\ldots \right) - 2 \left( \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots \right) = \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) - 2 \left( \sum_{n=1}^{\infty} \frac{1}{2n} \right)$

Step 3: We use the distributive property.

$\left( \sum_{n=1}^{\infty} \frac{1}{n} \right) - 2 \left( \sum_{n=1}^{\infty} \frac{1}{2n} \right) = \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) - \left( \sum_{n=1}^{\infty} \frac{2}{2n} \right) = \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) - \left( \sum_{n=1}^{\infty} \frac{1}{n} \right)$

Step 4: We combine the two sums.

$\left( \sum_{n=1}^{\infty} \frac{1}{n} \right) - \left( \sum_{n=1}^{\infty} \frac{1}{n} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n} \right) = \sum_{n=1}^{\infty} 0 = \boxed{0}$

Which step(s) is/are flawed, causing this proof to be invalid?

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