# Sequence and series

#### Elize

How can I solve this problem easily? please explain step by step

#### v8archie

Math Team
1. Decompose the fraction by the method of partial fractions.
2. Spot the telescoping series and thus remove many terms of the sum.
3. Sum the remaining terms.

idontknow

#### idontknow

Substitute for $$\displaystyle 2k=t \;$$ , the sum becomes $$\displaystyle s=\sum_{t=2}^{60} \dfrac{2}{t^2 -1 }$$.

Solve the equation for $$\displaystyle p \:$$ : $$\displaystyle \dfrac{2}{t^2 -1 }=p[\dfrac{1}{t-1}-\dfrac{1}{t+1} ] \; \Rightarrow$$ $$\displaystyle p=1$$.

Now the expression inside the sum(general term) becomes $$\displaystyle \dfrac{1}{t-1}-\dfrac{1}{t+1}$$.

$$\displaystyle s =\displaystyle \sum_{t=2}^{60} [\dfrac{1}{t-1}-\dfrac{1}{t+1}]=\sum_{t=2}^{60}\dfrac{1}{t-1} -\sum_{t=2}^{60}\dfrac{1}{t+1}$$.

$$\displaystyle s=[1+1/2+1/3+\dotsc +1/59]-[1/3+1/4+1/5+\dotsc +1/59 +1/60+1/61]=1+1/2-1/60-1/61$$.

#### skeeter

Math Team
If $2k=t$, then $t$ cannot be an odd value ...

$$\displaystyle \sum_{k=1}^{30} \dfrac{1}{2k-1} - \dfrac{1}{2k+1} = \dfrac{60}{61}$$

idontknow

#### idontknow

Correct version : $$\displaystyle \sum_{t=2k}^{60} 1/(t-1) - 1/(t+1)=[1+1/3+1/5+...+1/59]-[1/3+1/5+...+1/61]=1-1/61=60/61$$.