Show that the sum of the series ∑n(p^n) from n=1 to infinity (where 0<p<1) is S=p/(1-p)^2
To prove this, we can use the formula for the sum of an infinite geometric series. The formula states that for a geometric series with a common ratio between -1 and 1, the sum of the series is given by:
S = a / (1 - r)
where 'a' is the first term and 'r' is the common ratio of the series.
In our case, the first term 'a' is p, and the common ratio 'r' is p. Therefore, the formula becomes:
S = p / (1 - p)
So, this is the expression for the sum of the series ∑n(p^n) from n=1 to infinity.
Now let's simplify this expression further:
Simplifying the denominator:
1 - p = 1 - p^2/p = (p^2 - p)/p = p(1 - p)/p = (1 - p)/p
Substituting in the simplified denominator:
S = p / [(1 - p)/p]
S = p * (p/(1 - p))
S = p^2 / (1 - p)
Now, we can further simplify this expression using algebraic manipulation:
S = p^2 / (1 - p)
S = (p * p) / (1 - p)
S = p / [(1 - p)/p]
S = p / (1 - p)^(-1)
S = p / (1 - p)^2
Therefore, we have now shown that the sum of the series ∑n(p^n) from n=1 to infinity, where 0 < p < 1, is equal to S = p / (1 - p)^2.