How do you determine if #Sigma (2n3)/(5n+6)# from #n=[0,oo)# converge and find the sums when they exist?
The sum diverges.
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To determine convergence of the series Σ(2n3)/(5n+6) from n=0 to infinity, you can use the ratio test.

Apply the ratio test: Limit as n approaches infinity of (a_(n+1))/(a_n). Where a_n = (2n3)/(5n+6).

Find a_(n+1): a_(n+1) = (2(n+1)3)/(5(n+1)+6).

Substitute a_n and a_(n+1) into the ratio test: Limit as n approaches infinity of (2(n+1)3)/(5(n+1)+6) / ((2n3)/(5n+6)).

Simplify the expression: Limit as n approaches infinity of (2n+23)/(5n+11) * (5n+6)/(2n3). = Limit as n approaches infinity of (2n1)(5n+6)/(5n+11)(2n3).

Take the limit: Using the properties of limits, the limit evaluates to 10/10 = 1.

Analyze the result: Since the limit is equal to 1, the ratio test is inconclusive.

Apply another convergence test: You can try using the alternating series test if appropriate, or another convergence test such as the integral test.
For finding the sum when it exists, if the series converges, you can evaluate the sum using appropriate methods such as partial fraction decomposition, geometric series sum formula, or other techniques depending on the nature of the series.
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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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