sum of series

Jan 2010
142
0
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 \(\displaystyle r^n\).

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

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Mar 2010
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Chicago
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 \(\displaystyle r^n\).

I don't understand the problem. please help me out! thanks!
\(\displaystyle \sum_{k=0}^nar^k=a\frac{1-r^{n+1}}{1-r}\)

\(\displaystyle \sum_{k=0}^\infty ar^k=\frac{a}{1-r}\)

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 \(\displaystyle s_\infty-S_n\).
 
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