# 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?