Question

Find the sum of series 4/(4n-3) (4n+1)

Answer 1

`\frac4{(4n-3)(4n+1)}=\frac1{4n-3}-\frac1{4n-1}`

Assuming the sum starts at
`n=1`, the
`N`th partial sum is


`\displaystyle\sum_(n=1)^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\cdots+\left(\frac1{4N-7}-\frac1{4N-3}\right)+\left(\frac1{4N-3}-\frac1{4N+1}\right)`

`\displaystyle\sum_(n=1)^N\left(\frac1{4n-3}-\frac1{4n-1}\right)=1-\frac1{4N-1}`

As
`N\to\infty`, you're left with simply 1.