Find the sum of the first n terms of the series 1+2(1+1/n) +3(1+1/n)^2 + 4(1+1/n)^3....... Using the agp ( arithmerico - geometrico progression ) sum formula?

2 Answers
Aug 30, 2017

S_n=n^2

Explanation:

Let us find S_m, the sum of the first m terms of the series, as I do not want to mix it up with n in the series. If desired we can later put m=n to find sum of the first n terms of the series.

We have

S_m=1+2(1+1/n)+3(1+1/n)^2+4(1+1/n)^3.......+m(1+1/n)^(m-1) .......(A)

multiplying each term by (1+1/n), we get

S_m(1+1/n)=1(1+1/n)+2(1+1/n)^2+3(1+1/n)^3+4(1+1/n)^4.......+(m-1)(1+1/n)^(m-1)+m(1+1/n)^m .......(B)

Subtracting (B) from (A), we get

-S_m/n=1+(1+1/n)+(1+1/n)^2+(1+1/n)^3+...+(1+1/n)^(m-1)-m(1+1/n)^m

= ((1+1/n)^m-1)/(1+1/n-1)-m(1+1/n)^m

or S_m=mn(1+1/n)^m-n^2((1+1/n)^m-1)

and if m=n

S_n=n^2(1+1/n)^n-n^2((1+1/n)^n-1)=n^2

Aug 30, 2017

See below.

Explanation:

Using the inequality

sum_(k=1)^n x_k ge (prod_(k=1)^n x_k)^(1/n) we have

2/(n(n+1))sum_(k=0)^(n-1)(k+1)x^k = sum_(k=0)^(n-1) mu_k x^k ge prod_(k=0)^(n-1) x^(kmu_k)

where mu_k =(2 (k+1))/(n(n+1)) and sum_(k=0)^(n-1) mu_k = 1

Now calling phi = sum_(k=0)^(n-1) k mu_k we have

sum_(k=0)^(n-1)(k+1)x^k ge (n(n+1))/2 x^phi now making x = 1+1/n we have

sum_(k=0)^(n-1)(k+1)(1+1/n)^k ge (n(n+1))/2 (1+1/n)^phi