How do you find the interval of convergence #Sigma x^(2n)/(n*5^n)# from #n=[1,oo)#?

Question

Charles Arocha · Accepted Answer

#-sqrt5lt=xltsqrt5#

Find the ratio #abs(a_(n+1)/a_n)# for the series #suma_n#:

#abs(a_(n+1)/a_n)=abs(x^(2(n+1))/((n+1)5^(n+1))((n(5^n))/x^(2n)))=abs(x^2/5(n/(n+1)))#

The series converges if #lim_(nrarroo)abs(a_(n+1)/a_n)<1#.

#lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(x^2/5(n/(n+1)))=x^2/5lim_(nrarroo)abs(n/(n+1))=x^2/5#

So the series is convergent for

#x^2/5<1#

#x^2-5<0#

#(x-sqrt5)(x+sqrt5)<0#

#-sqrt5 < x < sqrt5#

Test both these endpoints by plugging them into the original series expression. Note that #x=sqrt5# gives the divergent series #sum_(n=1)^oo 1/n#, whereas #x=-sqrt5# gives the conditionally convergent series #sum_(n=1)^oo(-1)^n/n#. Thus, #x=-sqrt5# is included in the interval:

#-sqrt5lt=xltsqrt5#

Emily Ammerman · Answer

undefined To find the interval of convergence for the series Σ(x^(2n))/(n*5^n) from n=1 to infinity, we can use the ratio test. The ratio test states that if lim|a(n+1)/a(n)| as n approaches infinity is less than 1, the series converges. If it's greater than 1, the series diverges. And if it equals 1, the test is inconclusive.

For convergence, |x^2/5| < 1
So, |x^2| < 5
Which gives -√5 < x < √5

Thus, the interval of convergence is -√5 < x < √5.

Evelyn Agard · Answer

undefined To find the interval of convergence for the series Σ x^(2n) / (n * 5^n) from n = 1 to infinity:

1. Apply the ratio test to determine the convergence behavior of the series.
2. Use the ratio test formula: lim (|a_(n+1) / a_n|) as n approaches infinity.
3. Compute the ratio of consecutive terms: a_(n+1) / a_n.
4. Simplify the expression and compute its limit as n approaches infinity.
5. Analyze the limit to determine convergence or divergence.

For the given series:

a_n = x^(2n) / (n * 5^n)

To compute the ratio of consecutive terms:

a_(n+1) = x^(2(n+1)) / ((n+1) * 5^(n+1))

Calculate the ratio: a_(n+1) / a_n:

a_(n+1) / a_n = [(x^(2(n+1)) / ((n+1) * 5^(n+1))) / (x^(2n) / (n * 5^n))]

= [(x^(2(n+1)) / ((n+1) * 5^(n+1))) * ((n * 5^n) / x^(2n))]

= [x^2 / 5] * [(n * 5^n) / ((n+1) * 5^(n+1))]

= [x^2 / 5] * [n / (n+1)]

Now, take the limit of this expression as n approaches infinity:

lim (|a_(n+1) / a_n|) = lim [|x^2 / 5| * |n / (n+1)|] as n approaches infinity

= |x^2 / 5| * lim (|n / (n+1)|) as n approaches infinity

= |x^2 / 5| * 1

= |x^2 / 5|

The series converges if |x^2 / 5| < 1.

Therefore, the interval of convergence is determined by solving the inequality:

|x^2 / 5| < 1

|x^2| < 5

-√5 < x < √5

Hence, the interval of convergence is (-√5, √5).