# sum of series

#### 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 $$\displaystyle r^n$$.

#### undefined

MHF Hall of Honor
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 $$\displaystyle s_\infty-S_n$$.

• Anemori