Home > Calculus > Determining the Radius and Interval of Convergence for a Power Series

How do you find the radius of convergence #Sigma (x^n)/(5^sqrtn)# from #n=[0,oo)#?

Answer 1

The radius of convergence of the series:

#sum_(n=0)^oo x^n/5^sqrt(n)#

is #R=1#

To find the radius of convergence we can apply the ratio test, stating that a necessary condition for a series #sum_(n=1)^oo a_n# to converge is that:

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

If #L < 1 # the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

#abs (a_(n+1)/a_n) = abs ( frac (x^(n+1)/5^sqrt(n+1)) (x^n/5^sqrt(n))) = abs(x) 5^sqrt(n)/5^(sqrt(n+1)) = abs(x) 5^((sqrt(n) - sqrt(n+1)))#

Now we have:

#sqrt(n) - sqrt(n+1) = (( sqrt(n) - sqrt(n+1)) ( sqrt(n) + sqrt(n+1)))/ (sqrt(n) + sqrt(n+1)) = (n-(n+1))/(sqrt(n) + sqrt(n+1)) = -1/(sqrt(n) + sqrt(n+1))#

and as a result:

#lim_(n->oo) (sqrt(n) - sqrt(n+1)) = lim_(n->oo) -1/(sqrt(n) + sqrt(n+1)) = 0#

#lim_(n->oo) 5^(sqrt(n) - sqrt(n+1)) = 1#

so that:

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)#

which means that the series is absolutely convergent for #abs(x) < 1# and divergent for #abs(x) >1#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.