Is there a way to get an explicit formula for the sum after $n$ places of this summation:
$$\frac{1}{4} + \frac{3}{16} + \frac{9}{64} + \frac{27}{256}... = \frac{1}{4} + \frac{3}{4^2} + \frac{3^2}{4^3} + \frac{3^3}{4^4}... $$
In other words:
$S_1$ would be $\frac{1}{4}$
$S_2$ would be $\frac{7}{16}$
$S_3$ would be $\frac{37}{64}$
$S_n$ would be ?
$\endgroup$ 53 Answers
$\begingroup$The partial sum of a geometric series is given by
$\sum_{k=0}^{n} r^{k}=\frac{1-r^{n+1}}{1-r}$
provided $r \neq 1$. If $r=1$, then the sum is just $n+1$.
To see this, start with
$(1-r)(1+r+r^{2}+\ldots + r^{n})=1 + (r-r) + (r^{2}-r^{2}) + \ldots + (r^{n}-r^{n})-r^{n+1}$.
This "telescoping sum" simplifies to
$(1-r)(1+r+r^{2}+\ldots + r^{n})=1 - r^{n+1}$.
Dividing through by $(1-r)$ (assuming that $r \neq 1$) gives
$1+r+r^{2}+\ldots + r^{n}=\frac{1 - r^{n+1}}{1-r}$.
Your sum isn't quite in this form since it starts off with a first term of $1/4$, but it's easy to factor out the $1/4$ and get a geometric series starting with $1$.
$\endgroup$ $\begingroup$Because the initial term is $\frac 14$ and we multiply by $\frac 34$ for each term in our sum, we have
$$S_n = \frac{1}{4}\sum_{k=0}^n \bigg(\frac{3}{4}\bigg)^k$$
We know that summations of the form $\sum_{k=0}^n r^n $ are equal to $\frac{1 - r^{n}}{1-r}$, and so substituting $r=\frac 34$ gives
$$S_n = 1 - (3/4)^{n}$$
$\endgroup$ 1 $\begingroup$HINT: $$ S_n=a\left(\sum_{k=1}^{\infty}q^k-\sum_{k=n+1}^{\infty}q^k\right) $$ and $a$, $q$ and the expression for the sum of infinite geometric series you know.
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