Sum the following geometric series to infinity: `(sqrt(2)+1)+1+(sqrt(2)-1)+oo` `1/2+1/(3^3)+1/(2^3)+1/(3^4)+1/(2^5)+1/(3^6)+oo`

Explanation

(i) The given series is a geometric series with first term
`a=sqrt2+1` and the common ratio
`r=1/(sqrt2+1)=(sqrt2-1)/((sqrt2+1)(sqrt2-1))=sqrt2-1`
Hence, the sum to infinity is given by
`S=a/(1-r)=(sqrt2+1)/(1-(sqrt2-1))=(sqrt2+1)/(2-sqrt2)`
`=(sqrt2+1)/(sqrt2(sqrt2-1))`
`=((sqrt2+1)^(2))/(sqrt2(sqrt2-1)(sqrt+1))`
`=(3+2sqrt2)/(sqrt2)=(4+3sqrt2)/2`(ii) We have,
`1/2+1/(3^(2))+1/2^(3)+1/3^(4)+1/2^(5)+1/3^(6)+..."to"oo`
`=(1/2+1/2^(3)+1/2^(5)+...)+(1/3^(2)+1/3^(4)+1/3^(6)+...)`
`=(((1//2))/(1-(1//2^(2))))+(((1//3^(2)))/(1-(1//3^(2))))`
`=2/3+1/8=19/24`