The problem:
Given that
$$\frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} – \frac{1}{7} – \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} – \frac{1}{15} + \ldots \\
\\
\frac{\pi}{12} & = \frac{1}{3} – \frac{1}{9} + \frac{1}{15} – \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} – \frac{1}{45} + \ldots\\
\\
& = 0 + \frac{1}{3} + 0 + 0 – \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 – \frac{1}{21}
\end{align}
$$
now add them together:
$$
\begin{align}
\frac{\pi}{4} + \frac{\pi}{12} & = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} – \frac{1}{11} + \frac{1}{13} – \frac{1}{15} + \ldots \\
\\
& + 0 + \frac{1}{3} + 0 + 0 – \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\
\end{align}
$$
and we will get:
$$
\begin{align}
\frac{\pi}{3} & = 1 + 0 + \frac{1}{5} – \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\
& = 1 + \frac{1}{5} – \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots
\end{align}
$$
My questions:
-
I inserted/removed infinite zeros into/from the series, is that OK?
-
My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)
- Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?
Best Answer