How do you determine the convergence or divergence of #Sigma ((-1)^(n)n)/(n^2+1)# from #[1,oo)#?
As:
and: the series is convergent.
An alternative series:
So we make the test:
As the denominator is always positive we can focus on the numerator:
#( (n+1)(n^2+1)-n( (n+1)^2+1)) = n^3+n+n^2+1 -n (n^2+2n+1+1) = n^3+n+n^2+1 -n^3-2n^2-2n = -2n^2-2n +1 <=0#
The we check that:
So both conditions are satisified and the series is convergent.
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Using this in conjunction with the fact that
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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.
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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