How do you find the interval of convergence Sigma 4^n/(3^n+5^n)x^n from n=[0,oo)?

1 Answer
Jan 16, 2017

The series:

sum_(n=0)^oo (4^n)/(3^n+5^n)x^n

is convergent in the interval x in (-5/4,5/4) where it is absolutely convergent.

Explanation:

The ratio test states that a series sum_(n=0)^oo a_n converges absolutely if:

lim_(n->oo) abs (a_(n+1)/a_n) < 1

Let us determine the ratio for the series:

sum_(n=0)^oo (4^n)/(3^n+5^n)x^n

abs (a_(n+1)/a_n) = (4^(n+1)/(3^(n+1)+5^(n+1))abs(x)^(n+1))/(4^n/(3^n+5^n)abs(x)^n)=4abs(x) (3^n+5^n)/(3^(n+1)+5^(n+1))=4/5abs(x) (1+(3/5)^n)/(1+(3/5)^(n+1))

Now, as 3/5 < 1 we have that:

lim_(n->oo) (3/5)^n = 0

so that:

lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) 4/5abs(x) (1+(3/5)^n)/(1+(3/5)^(n+1)) = 4/5absx

We can then conclude that for:

absx <5/4 => lim_(n->oo) abs (a_(n+1)/a_n) <1 and the series is absolutely convergent

absx > 5/4 => lim_(n->oo) abs (a_(n+1)/a_n) >1 and the series is divergent,

The case where abs(x) = 5/4 is indeterminate and we have to analyze in detail.

In the case where x=+-5/4 we have:

abs(a_n) = 4^n/(3^n+5^n)(5/4)^n = 5^n/(3^n+5^n) = 1/(1+(3/5)^n)

so that:

lim_(n->oo) abs(a_n) = 1 >0

and that means that the series cannot converge.

In conclusion the series is convergent in the interval x in (-5/4,5/4) where it is also absolutely convergent.