Sequence summation GRE question 
This problem is giving me a lot of trouble. the only way i can think of doing it is to take the integeral a(k) from 1 to 100 but that is definitely not what i am supposed to do since its for the GRE and no knowledge of calculus is necessary. . . helpppp
 A: It is a telescopic sum, note for example that for $k=1,2,3$ then 
$$ a_1 + a_2 + a_3 = (1-1/2)+(1/2-1/3)+(1/3-1/4) = 1 - 1/4$$
so the intermediate values always cancel.
Then you can easily check that
$$ \sum_{k=1}^{100} a_k = 1-1/101=100/101$$
A: HINT:  Try writing out the first few terms...do things seem to cancel out?  Can you figure out what is left at the end?
A: $a_1 = \frac{1}{1} - \frac{1}{2} $
$a_2 = \;\;\;\;\;\;\frac{1}{2} - \frac{1}{3} $
$a_3 = \;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{3} - \frac{1}{4}$  
$\vdots \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ddots\ddots$
$a_{n-1} = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \frac{1}{n-1}-\frac{1}{n}$
$a_{n} = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\; \;\;\;\frac{1}{n}-\frac{1}{n+1}$
Now add them all up and you get? 
A: $a_1=1-\frac{1}{2}$, $a_2=\frac{1}{2}-\frac{1}{3}$, ... , $a_{100}=\frac{1}{100}-\frac{1}{101}$
So the sum is $1-\frac{1}{101} = \frac{100}{101}$ (Other terms cancel out. Only the first and last terms are left.)
A: If you write out the first few terms of the series, you'll see that $$a_1+a_2+a_3+\cdots+a_{99}+a_{100}$$$$=\left(1-\frac{1}{2}\right)+ \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3} - \frac{1}{4} \right) \cdots+\left(\frac{1}{99} - \frac{1}{100}\right) + \left( \frac{1}{100} - \frac{1}{101} \right).$$
Rearranging the parentheses yields $$1 + \left(-\frac{1}{2} + \frac{1}{2}\right) + \left(-\frac{1}{3} + \frac{1}{3}\right) + \cdots +\left(-\frac{1}{100} + \frac{1}{100} \right) - \frac{1}{101}$$ $$=1-\frac{1}{101} = \frac{100}{101}.$$

You can confirm that the general formula for the $n$th sum is $$\sum_{k=1}^{n}\frac{1}{k} - \frac{1}{k+1}=1-\frac{1}{n+1}$$ by induction, since it is clearly true for $n=1$, and then assuming that it holds for $n$, we can see that $$\sum_{k=1}^{n+1}\frac{1}{k} - \frac{1}{k+1}=\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{n+1} - \frac{1}{n+2} \right)$$
so from our inductive hypothesis we get
$$\left(1-\frac{1}{n+1}\right)+\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$
$$=1-\frac{1}{n+2}.$$
A: It is easiest to see the answer when you write the sum in two parts
$\sum_{k=1}^{100} \frac{1}{k} -\sum_{k=1}^{100} \frac{1}{k+1} $  
$= (1+\frac{1}{2}+\frac{1}{3}+ \dots + \frac{1}{100})-(\frac{1}{2}+\frac{1}{3}+ \dots +\frac{1}{101})$
