Question

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFraction plus StartFraction (negative 1 )Superscript n Over 5 Superscript n EndFraction ). What is the​ series' sum?

Answer 1

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series  

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$      and that equality is true as long as     $|r| < 1$.

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$        and   $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$

Therefore we can use the formula and

$\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n = \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16 = 15/8$