How do you know what $\sum_{n=a}^{\infty}p^n$ if you know $\sum_{n=0}^{\infty}p^n = \frac{1}{1-p}$ ?
Apparently $\sum_{n=a}^{\infty}p^n$ = $\frac{p^a}{1-p}$, but how can you derive this?
How do you know what $\sum_{n=a}^{\infty}p^n$ converges to if you know $\sum_{n=0}^{\infty}p^n = \frac{1}{1-p}$?
2
$\begingroup$
sequences-and-series
geometric-series
-
3Write it as $p^a\sum_{n=0}^\infty p^n$ – 2017-02-24
-
2Observe: $\sum_{n=a}^\infty p^n=p^a\sum_{n=0}^\infty p^n$. – 2017-02-24
-
1The difference between the two is $\sum_{n=0}^{a-1}p^n$. – 2017-02-24
2 Answers
3
$$ \begin{align} \sum_{n=a}^\infty p^n &= \sum_{n=0}^\infty p^{n+a} \\ &= p^a \sum_{n=0}^\infty p^{n} \\ &= \frac{p^a}{1-p} \end{align} $$
1
Because of the high-school identity: $$1-p^n=(1-p)(1+p+p^2+\dots+p^{n-1}),$$ rewritten as $$\frac1{1-p}=1+p+p^2+\dots+p^{n-1}+\frac{p^n}{1-p}, $$ which shows that $$\Biggl\lvert\frac1{1-p}-(1+p+p^2+\dots+p^{n-1})\Biggr\rvert=\frac{\lvert p^n\rvert}{\rvert 1-p\rvert},$$ and the latter fraction tends to $0$ if $\lvert p\rvert<1$.