# Apply property in convergent series and show series u=3^n-5^n/15^n, n> or equal to 1, convergent. Calculate sum?

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You need to write the series such that: `u_n =(3^n-5^n)/(15^n) = u=(3^n)/(15^n)-(5^n)/(15^n)`

Reducing yields:

`u_n = 1/(5^n) - 1/(3^n) = (1/5)^n - (1/3)^n`

Notice that series `(1/5)^n` and `(1/3)^n` converge.

You need to find the sum of series u, hence you need to find the sum of each series such that:

`lim _(n-gtoo) sum_(n=1)^oo (1/5)^n =lim _(n-gtoo) (1/5 + (1/5)^2 + ... + (1/5)^n) = (1/5)*lim _(n-gtoo)(1- (1/5)^n)/(1 - 1/5) = (1/5)/(4/5) =gt lim _(n-gtoo) sum_(n=1)^oo (1/5)^n = 1/4`

`lim _(n-gtoo) sum_(n=1)^oo (1/3)^n = (1/3)*lim _(n-gtoo)(1- (1/3)^n)/(1 - 1/3) = (1/3)/(2/3) =gt lim _(n-gtoo) sum_(n=1)^oo (1/3)^n = 1/2`

`` Hence `sum_(n=1)^oo u_n = sum_(n=1)^oo (1/5)^n - sum_(n=1)^oo (1/3)^n = 1/4 - 1/2 = -1/4`

**Hence, the series `u_n=(3^n-5^n)/(15^n)` converges and evaluating its sum yields `sum_(n=1)^oo u_n = -1/4` .**