How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #1+3/4+9/16+...+(3/4)^n+...#?

Question

Ezekiel Alexander · Accepted Answer

undefined To find the nth partial sum of the series $1 + \frac{3}{4} + \frac{9}{16} + \ldots + \left(\frac{3}{4}\right)^n + \ldots$, you use the formula for the sum of a geometric series. The formula is $S_n = \frac{a(1 - r^n)}{1 - r}$, where $S_n$ is the nth partial sum, $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

For this series, the first term $a$ is 1, and the common ratio $r$ is $ \frac{3}{4} $. Plug these values into the formula to find the nth partial sum.

To determine whether the series converges, you check if the absolute value of the common ratio $r$ is less than 1. If $|r| < 1$, the series converges; otherwise, it diverges.

If the series converges, you use the formula for the sum of an infinite geometric series, which is $S = \frac{a}{1 - r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.

Plug in the values of $a$ and $r$ into the formula to find the sum of the series if it exists.

Ezra Adams · Answer

#sum_(k=0)^n (3/4)^k = 4-3(3/4)^(n-1)#

#sum_(k=0)^oo (3/4)^k = 4# This is a geometric series of ratio #q = 3/4#. Consider the series: #sum_(k=0)^oo q^k# and its partial sum: #s_n = sum_(k=0)^n q^k = 1+q+q^2+...+q^n = (q^(n+1)-1)/(q-1)# Then, if #abs q < 1#: #lim_(n->oo) s_n = lim_(n->oo) (q^(n+1)-1)/(q-1) = 1/(1-q)# For #q=3/4#: #s_n = ((3/4)^n-1)/(3/4-1) = (3^n-4^n)/(4^n(-1/4))=(4^n-3^n)/4^(n-1) = 4-3(3/4)^(n-1)# and: #sum_(k=0)^oo (3/4)^k = 4#