What is the summation i=1 to n of (i ^ -1)?

Answer 1

i^(-1) = 1/i = -i

summation from i = 1 to n i^(-1) = -ni

put if you mistyped and meant which is likely

i^(-n) it is geometic series

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

Answer 2

If the summation is i^-1, then it would just be the same as 1/i, which is 1, no matter what n is. I don't understand: where is the n in the summation?

Answer 3

The sum [k=1, n]∑1/k is called the nth harmonic number. It is given by [0, 1]∫(1-x^n)/(1-x) dx. This can be shown by induction:

First, this formula holds where n=1, since:
[k=1, 1]∑1/k = 1, and
[0, 1]∫(1-x)/(1-x) dx = [0, 1]∫1 dx = 1

Now, suppose this formula holds for some n. Then:

[k=1, n+1]∑1/k
[k=1, n]∑1/k + 1/(n+1)

By our inductive hypothesis:

[0, 1]∫(1-x^n)/(1-x) dx + 1/(n+1)

The last term is equal to [0, 1]∫x^n dx, so:

[0, 1]∫(1-x^n)/(1-x) dx + [0, 1]∫x^n dx
[0, 1]∫(1-x^n)/(1-x) + x^n dx
[0, 1]∫(1-x^n)/(1-x) + (x^n - x^(n+1))/(1-x) dx
[0, 1]∫(1-x^(n+1))/(1-x) dx

Thus the formula holds for n+1, and by induction, it holds for all n.

Answer 4

this is 1+1/2 +1/3 + 1/4 +..., right?

This is the harmonic series. To sum, let Sn be the sum to n
let S(n+1) be Sn +1/(n+1)

Hmm, I forgot!