# How do you test the series #Sigma 2/(4n-3)# from n is #[1,oo)# for convergence?

To test the series Σ 2/(4n - 3) from n = 1 to infinity for convergence, you can use the Ratio Test.

First, calculate the limit of the absolute value of the ratio of the (n+1)th term to the nth term as n approaches infinity:

lim (n→∞) |(a_{n+1}) / (a_n)|

If the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit doesn't exist, the series diverges. If the limit equals 1, the test is inconclusive.

For the given series:

a_n = 2/(4n - 3)

a_{n+1} = 2/(4(n+1) - 3) = 2/(4n + 1)

Then, calculate the limit:

lim (n→∞) |(a_{n+1}) / (a_n)| = lim (n→∞) |(2/(4n + 1)) / (2/(4n - 3))| = lim (n→∞) |(4n - 3) / (4n + 1)| = 1

Since the limit equals 1, the Ratio Test is inconclusive. Inconclusiveness in the Ratio Test suggests that further tests, like the Limit Comparison Test or another method, may be needed to determine convergence or divergence.

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We can apply d'Alembert's ratio test:

Suppose that;

Then

if L < 1 then the series converges absolutely; if L > 1 then the series is divergent; if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

So our test limit is:

and we can conclude that in the case the ratio test is inconclusive, (in fact the series diverges)

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