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

So the series is convergent for

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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.

Using the ratio test:
lim |[(x^(2(n+1)))/(n+1)*(5^(n))]/[(x^(2n))/(n*5^n)]| as n approaches infinity
= lim |(x^2*(n)/(n+1)) / 5| as n approaches infinity
= |x^2/5|

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

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

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

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

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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.

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.

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.

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