Question

How do you find the sum of the infinite series `Sigma(1/10)^k` from k=1 to `oo`?

Answer 1

`color(blue)(1/9)`

Explanation:

`sum _(k=1)^(oo)(1/10)^k`

Find the common ratio, by calculating the first three terms.

`(1/10)^1, (1/10)^2, (1/10)^3= 1/10, 1/100, 1/1000`

Ratio:

`(1/100)/(1/10)=(1/1000)/(1/100)=1/10`

This could have been seen from the summation expression.

The sum of a geometric series is:

`a((1-r^n)/(1-r))`

Where `a` is the first term, `r` is the common ratio and `n` is the nth term.

So:

`1/10((1-0)/(1-(1/10)))=1/10(1/(1/(9/10)))=1/10(10/9)=color(blue)(1/9)`

This could also have been arrived at using the limit to `oo`

`1/10lim_(n->oo)((1-(1/10)^n)/(1-(1/10)))=10/9=1/10*10/9=1/9`