# Show how calculate summation ((2n+5)/(3n-10))^n, if n go to infinit?

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You need to use the Cauchy's rule such that if`lim_(n-gtoo) (a_(n+1))/(a_n) = k, klt1` , then the series sum `a_n` converges.

Hence, considering `sum_(n=1)^oo((2n+5)/(3n-10))^n, then a_n =((2n+5)/(3n-10))^n`

Using Cauchy's rule yields:

`root(n)((2n+5)/(3n-10))^n= (2n+5)/(3n-10)`

Evaluating the limit of `a_n` yields:

`lim_(n-gtoo) (2n+5)/(3n-10) = 2/3` (notice that the orders of numerator and denominator matches, hence the limit is the ratio of leading coefficients)

Since `2/3 lt 1` => series `sum_(n=1)^oo((2n+5)/(3n-10))^n` .

**Hence, using Cauchy's rule yields that series `sum_(n=1)^oo((2n+5)/(3n-10))^n` converges.**