1. ## sum of series

i hope this is the right place:

Find the sum of:
$\displaystyle \sum_{n=1} ^{10} {3*(\frac{1}{4})^n}$

$\displaystyle =1-(\frac{1}{4})^{10}) = 1$

but there is an additional problem:

Find the error between $\displaystyle S_{10}$ and $\displaystyle s_\infty$. Use the formula for $\displaystyle S_n$ and $\displaystyle s_\infty$ of convergement geometric series , find a general formula that gives the error between $\displaystyle S_n$ and $\displaystyle s_\infty$. Use the new formula to explain why, for the series in #7, the error between $\displaystyle S_n$ and $\displaystyle s_\infty$ is always equal to $\displaystyle r^n$.

2. Originally Posted by Anemori
i hope this is the right place:

Find the sum of:
$\displaystyle \sum_{n=1} ^{10} {3*(\frac{1}{4})^n}$

$\displaystyle =1-(\frac{1}{4})^{10}) = 1$

but there is an additional problem:

Find the error between $\displaystyle S_{10}$ and $\displaystyle s_\infty$. Use the formula for $\displaystyle S_n$ and $\displaystyle s_\infty$ of convergement geometric series , find a general formula that gives the error between $\displaystyle S_n$ and $\displaystyle s_\infty$. Use the new formula to explain why, for the series in #7, the error between $\displaystyle S_n$ and $\displaystyle s_\infty$ is always equal to $\displaystyle r^n$.

$\displaystyle \sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r}$
$\displaystyle \sum_{k=0}^\infty ar^k=\frac{a}{1-r}$
Since we are dealing with sums starting with $\displaystyle k=1$, we'll need to subtract that out.
$\displaystyle S_n=\sum_{k=1}^nar^k=a\frac{1-r^{n+1}}{1-r}-a$
$\displaystyle s_\infty = \sum_{k=1}^\infty ar^k=\frac{a}{1-r}-a$
The term error is used because the sequence $\displaystyle S_1, S_2, S_3, \cdots$ can be seen as approximations to $\displaystyle s_\infty$ that get closer and closer as $\displaystyle n$ increases. The error is the true value minus the value of the approximation, or $\displaystyle s_\infty-S_n$.