sum of series

**Anemori** · May 18th 2010, 09:22 PM

i hope this is the right place:

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

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

but there is an additional problem:

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

I don't understand the problem. please help me out! thanks!

**undefined** · May 18th 2010, 09:34 PM

Originally Posted by Anemori

i hope this is the right place:

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

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

but there is an additional problem:

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

I don't understand the problem. please help me out! thanks!

$\sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r}$

$\sum_{k=0}^\infty ar^k=\frac{a}{1-r}$

Since we are dealing with sums starting with $k=1$ , we'll need to subtract that out.

$S_n=\sum_{k=1}^nar^k=a\frac{1-r^{n+1}}{1-r}-a$

$s_\infty = \sum_{k=1}^\infty ar^k=\frac{a}{1-r}-a$

The term error is used because the sequence $S_1, S_2, S_3, \cdots$ can be seen as approximations to $s_\infty$ that get closer and closer as $n$ increases. The error is the true value minus the value of the approximation, or $s_\infty-S_n$ .

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