Basic Geometric Series Question-Stuck

Question

I'm studying Calc 2 and I have a basic series question.

A geometric series $\sum _{n=0}^{\infty }\:Ar^n$ is convergent if |r|<1 and the sum equals $\frac{a}{1-r}$ if the series is convergent.

Question: $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$

and $\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ totals $\frac{5}{1-\frac{1}{\pi }}=\frac{5\pi \:}{\pi \:-1}$.

Therefore, $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ = $\frac{5\pi \:}{\pi \:-1}\:-\frac{5}{\pi }$.

I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.

Answer 1

I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.

$-\dfrac5{\pi^0}=-5$, because $\pi^0=1$.

Answer 2

You did almost everything right. You took a sum from 1 to $\infty$ and added a $n=0$ term, and then subtracted it (your $\frac{-5}{\pi}$ term).

But you wrote it wrong, because $$ \frac{5}{\pi^0} = \frac{5}{1} = 5. $$

Nice work otherwise!

There's another way to deal with a series starting at $1$. You write \begin{align} S &= \sum_{n=1}^\infty \frac5{\pi^n} \\ \end{align} and then you let $k = n-1$ in that expression, or $n = k + 1$ (which si the same thing), and rewrite the sum in terms of $k$. As $n$ goes from $1$ to $\infty$, $k$ goes from $0$ to $\infty$. So you get \begin{align} S &= \sum_{n=1}^\infty \frac5{\pi^n} \\ &= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\ \end{align} Now that's not actually in geometric-series form, but a little algebra -- $\pi^{k+1} = \pi \cdot \pi^k$, is enough to get you there: \begin{align} S &= \sum_{n=1}^\infty \frac5{\pi^n} \\ &= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\ &= \sum_{k=0}^\infty \frac5{\pi\cdot\pi^{k+1}} \\ &= \sum_{k=0}^\infty \frac5{\pi}\cdot\frac{1}{\pi^{k}} \\ &= \frac{5}{\pi}\sum_{k=0}^\infty \frac{1}{\pi^{k}} \\ &= \frac{5}{\pi}\sum_{k=0}^\infty \left(\frac{1}{\pi}\right)^{k} \\ \end{align} and that's a geometric series with $A = 1$ and $r = \frac1\pi$.