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

`\sum\limits_(n=0)^(\infty) (2)/(3^n)((-1)^n)/(5^n) = 15/8`

Explanation:

The sum you are trying to understand is this.

`\sum\limits_(n=0)^(\infty) (2)/(3^n)((-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 = (a)/(1-r)` and that equality is true as long as
`|r| < 1`.

Therefore here we have

`\sum\limits_(n=0)^(\infty) (2)/(3^n)((-1)^n)/(5^n) = \sum\limits_(n=0)^(\infty) 2* \big((-1)/(3*5) \big)^n = \sum\limits_(n=0)^(\infty) 2* \big((-1)/(15) \big)^n` and
`\big|(-1)/(15) \big| = (1)/(15) < 1`

Therefore we can use the formula and

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