Question

Find a formula for the nth partial sum of the series and use it to find the​ series' sum if the series converges. StartFraction 13 Over 2 times 3 EndFraction plus StartFraction 13 Over 3 times 4 EndFraction plus StartFraction 13 Over 4 times 5 EndFraction plus midline ellipsis plus StartFraction 13 Over (n plus 1 )(n plus 2 )EndFraction plus midline ellipsis

Answer 1

Looks like the series is supposed to be

`(13)/(2*3)+(13)/(3*4)+(13)/(4*5)+\cdots+(13)/((n+1)(n+2))`

or in condensed form,

`\displaystyle\sum_(k=1)^n(13)/((k+1)(k+2))`

Decompose the summand into partial fractions:

`(13)/((k+1)(k+2))=\frac a{k+1}+\frac b{k+2}`

`\implies 13=a(k+2)+b(k+1)`

`k=-1\implies13=a`

`k=-2\implies13=-b\implies b=-13`

So the
`n`th partial sum of the series is

`S_n=\displaystyle\sum_(k=1)^n(13)/(k+1)-(13)/(k+2)`

`\displaystyle S_k=\left(\frac{13}2-\frac{13}3\right)+\left(\frac{13}3-\frac{13}4\right)+\cdots+\left(\frac{13}n-(13)/(n+1)\right)+\left((13)/(n+1)-(13)/(n+2)\right)`

`\implies S_k=\frac{13}2-(13)/(n+2)`

As
`n\to\infty`, the second term converges to 0, leaving the value of the infinite series,

`\displaystyle\sum_(k=1)^\infty(13)/((k+1)(k+2))=\boxed{\frac{13}2}`