How do you test the series #Sigma (n+3)/(n(n+1)(n2))# from n is #[3,oo)# for convergence?
To test the convergence of the series Sigma (n+3)/(n(n+1)(n2)) from n is [3,oo), you can use the Ratio Test. Let's denote the general term of the series as a_n = (n+3)/(n(n+1)(n2)).

Apply the Ratio Test:
 Compute the limit as n approaches infinity of the absolute value of the ratio of consecutive terms: lim (n→∞) a_(n+1)/a_n.

Simplify the expression:
 Substitute (n+1) for n in the general term to get a_(n+1), then divide a_(n+1) by a_n.

Take the limit:
 Evaluate the limit of the ratio as n approaches infinity.

Analyze the result:
 If the limit is less than 1, the series converges absolutely.
 If the limit is greater than 1 or undefined, the series diverges.
 If the limit equals 1, the test is inconclusive, and another test may be needed.

If the limit is less than 1, conclude that the series converges absolutely. If the limit is greater than 1 or undefined, conclude that the series diverges.
Apply these steps to determine the convergence of the given series.
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is convergent.
Given:
We can start from:
if we increase the numerator and decrease the denominator, the resulting quotient will be bigger:
is convergent.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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