How do you test for convergence of #Sigma (3n-7)/(10n+9)# from #n=[0,oo)#?

Answer 1

By putting an limit in front.

#lim_(n=>oo)((3n-7)/(10n+9))# In this case, if the result is #!=0# there's divergence. But the result is something that we cannot solve. #=>[oo/oo]# So we are going to isolate n in the equation: #=>lim_(n=>oo)((n(3-7/n))/(n(10+9/n)))# #=lim_(n=>oo)((3-7/n)/(10+9/n))# And we just have to solve it! #=(3-7/oo)/(10+9/oo)# #=(3-0)/(10+0)# #=3/10# The result is #!=0# so there's divergence! (I'not an english speaker so I don't know the name of this method. Feel free to add it.)
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Answer 2

To test the convergence of the series Σ(3n-7)/(10n+9) from n=0 to infinity, you can use the limit comparison test. Let's compare it with the series Σ1/n.

First, find the limit of the ratio of the two series as n approaches infinity:

lim (3n-7)/(10n+9) / (1/n) = lim (3n-7)/(10n+9) * n = lim (3n^2 - 7n) / (10n + 9) = lim (3 - 7/n) / (10 + 9/n) = 3/10

Since the limit is a finite nonzero number, and 1/n is a p-series with p=1, which is known to converge, by the limit comparison test, the given series also converges.

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Answer from HIX Tutor

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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