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

Question

Scarlett Allison · Accepted Answer

The interval of convergence is #-1<=x<=1#

Use the Ratio test to compute the convergence interval

#a_n=x^n/n^2#

#|a_(n+1)/a_n|=|x^(n+1)/(n+1)^2*n^2/x^n|#

#=(n^2x)/(n+1)^2#

Calculate the limit as #n->oo#

Therefore,

#lim_(n->oo)|(n^2x)/(n+1)^2|#

#=|x|*lim_(n->oo)|(n^2)/(n+1)^2|#

#=|x|*1#

#=|x|#

The series converges for #|x|<1#

Now, check the convergence when, #|x|=1#

When #x=1#

#sum_1^oo1^n/n^2#

By the #" p-series test"#, #p=2# and #p>1#, the series converges

When #x=-1#

#sum_1^oo(-1)^n/n^2=sum_1^oo|(-1)^n/n^2|=sum_1^oo1/n^2#

By the #" p-series test"#, #p=2# and #p>1#, the series converges

The series converges absolutely

Elijah Abram · Answer

undefined To find the interval of convergence for the series Σ(x^n/n^2) from n=1 to infinity, you can use the ratio test.

First, apply the ratio test:
- Take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term.

The series converges if the limit is less than 1 and diverges if the limit is greater than 1.

In this case, the series is Σ(x^n/n^2). So, applying the ratio test:
- Take the absolute value of the ratio of the (n+1)th term to the nth term, which gives ((x^(n+1))/(n+1)^2) / ((x^n)/n^2).
- Simplify the expression to get ((x^(n+1))(n^2))/((x^n)((n+1)^2)).
- Cancel out the common terms to get x/(1 + 2/n), which simplifies to x as n approaches infinity.

Since the limit of x as n approaches infinity is x, the series will converge for |x| < 1.

Therefore, the interval of convergence is -1 < x < 1.