G.P. and Infinite Series Problem How many terms of the geometric series 7, 7/2, 7/4, 7/8, ... must be taken in order that the sum may differ from the sum to infinity by less than 0.01? (Ans: 11 terms)
Some help, please? 
 A: Hint: Let $x=1/2$ in the formula
$$
1 + x +x^2 + \cdots + x^n = \frac{{1 - x^{n + 1} }}{{1 - x}}.
$$
EDIT:
Let $S_n$ be the $n$th partial sum, so that
$$
S_1 = 7,
$$
$$
S_2 = 7 + 7/2 = 7(1+1/2),
$$
$$
S_3 = 7 + 7/2 + 7/4 = 7(1+1/2+1/4),
$$
from which you can conclude that
$$
S_n = 7 (1+1/2+\cdots+1/2^{n-1}).
$$
Further, let $S$ denote the sum to infinity, so that
$$
S = 7 (1+1/2+1/4+\cdots) = 7 \cdot 2 = 14.
$$
You are asked to find the least $n$ such that
$$
S - S_n < 0.01.
$$
Noting that
$$
S_n = 7 (1+1/2+\cdots+1/2^{n-1}) = 7 \frac{{1 - (1/2)^n }}{{1 - 1/2}} = 14(1 - 1/2^n ),
$$
it remains to find the least $n$ such that
$$
14 - 14(1 - 1/2^n ) < 0.01,
$$
or
$$
14/2^n < 0.01,
$$
or
$$
2^n > 1400.
$$
Since $2^{10} = 1024 < 1400$ and $2^{11} = 2048 > 1400$, the answer is $11$.
A: Do you know how to sum a geometric series?  Basically you are being asked how late to start so that the sum from there on is less than 0.01.  So if you write the sum starting from $n$ to $\infty$ you should be able to find $n$ that meets this.  A geometric series stays geometric if you cut off some of the first terms.
A: HINT:
We know how to sum geometric series ($\dfrac{a_0}{1-r}$ where $a_0$ is the first term and $r$ is the ratio). We also know how to sum partial geometric series, by either directly adding them up (applicable here) or by recalling some useful formulas. 
You could subtract these, or you could realize that the difference is just another geometric series with a different starting point.
