Question

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?

Answer 1

`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`

Answer 2

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`