Let `S_(n)=1+(1)/(2)+(1)/(3)+….+(1)/(n)` Show that `s_(n)=n-((1)/(2)+(2)/(3))+…….+(n-1)/(n))`.

Let S_(n)=1+(1)/(2)+(1)/(3)+….+(1)/(n) Show that ns_(n)=n+((n-1)/(1)+(n-2)/(2)+………+(2)/(n-2)+(1)/(n-1)) .

Let S_(n)=1+(1)/(2)+(1)/(3)+(1)/(4)+......+(1)/(2^(n)-1) Then

If S_(n)=1 + (1)/(2) + (1)/(2) + …..+ (1)/(2^(n-1)), (n in N) then …….

If H_(n) =1+(1)/(2)+ cdots + (1)/(n) then value of S_(n) =1 + (3)/(2) + (5)/(3) +cdots + (2n-1)/(n) is

If S_(1), S_(2), S_(3),…., S_(n) are the sums to infinity of n infinte geometric series whose first terms are 1,2,3,… n and whose common ratios are (1)/(2), (1)/(3), (1)/(4), ….(1)/(n+1) respectively, show that, S_(1) + S_(2) + S_(3) +…S_(n) = (1)/(2)n(n+3)

If S= tan^(-1) ((1)/(n^(2) +n+1)) + tan^(-1) ((1)/(n^(2) + 2n+ 3))+ ….+ tan^(-1) ((1)/(1+(n+19) (n+20))) , then tan (s) is equal to

Let S_n=1+1/2+1/3+1/4+......+1/(2^n-1). Then

Suppose a series of n terms given by S_(n)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n) then S_(n-1)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n-1),nge1 subtracting we get S_(n)-S_(n-1)=t_(n),nge2 surther if we put n=1 is the first sum then S_(1)=t_(1) thus w can write t_(n)=S_(n)-S_(n-1),nge2 and t_(1)=S_(1) Q. The sum of n terms of a series is a.2^(n)-b . where a and b are constant then the series is

Suppose a series of n terms given by S_(n)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n) then S_(n-1)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n-1),nge1 subtracting we get S_(n)-S_(n-1)=t_(n),nge2 surther if we put n=1 is the first sum then S_(1)=t_(1) thus w can write t_(n)=S_(n)-S_(n-1),nge2 and t_(1)=S_(1) Q. The sum of n terms of a series is a.2^(n)-b . where a and b are constant then the series is

If a_(1),a_(2),a_(3)….a_(n) are positive and (n-1)s = a_(1)+a_(2)+….+ a_(n) then prove that a_(1),a_(2),a_(3)…a_(n) ge (n-1)^(n)(s-a_(1))(s-a_(2))….(s-a_(n))