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In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?

The answer given is $\frac{100}{101}$, but I am not sure how.

So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$

The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator.

  • 0
    Have you heard of telescoping sums? – 2017-02-08
  • 0
    Write out the sum for a few terms and notice the cancellations. – 2017-02-08
  • 0
    What can you say about $$ (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) $$ – 2017-02-08
  • 0
    Try writing $a_{k+1}+a_k$ then add $a_{k+2}$ to that and try to see a pattern. – 2017-02-08

3 Answers 3

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$$a_1+a_2+a_3+...+a_{100}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)+\left(\frac{1}{100}-\frac{1}{101}\right)=1-\frac{1}{101}=\frac{100}{101}$$

Note that $-\frac{1}{2}$ cancel $\frac{1}{2}$, $-\frac{1}{3}$ cancel $\frac{1}{3}$ and go on until $-\frac{1}{100}$ cancel $\frac{1}{100}$.

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$$a_{100} = \left( \frac11 - \frac12 \right)+\left( \frac12 - \frac13 \right)+ \left( \frac13 - \frac14 \right)+\cdots + \left( \frac1{100} - \frac1{101} \right) \\= \frac11 + \left( -\frac12+\frac12\right)+\left( -\frac13+\frac13\right) +\cdots +\left( -\frac1{100}+\frac1{100}\right) -\frac1{101} = 1 - \frac1{101} $$

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Such a sum is called telescop sum. We have $$\sum_{k=1}^{100} \left( \frac{1}{k}- \frac{1}{k+1} \right) \stackrel{\star}{=}1 - \frac{1}{101}.$$ In $(\star)$ we split the sum into two and make an index shift.