Series Convergence Test for ∑ 5^n/(4^n +3)

bigu01
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Homework Statement


Which of the series, diverge or converge ∑ 5^n/(4^n +3 )


Homework Equations





The Attempt at a Solution

Taking the limit as n→∞ we have (5^n ln 5)/ (4^n ln 4) , my question is here how does it become like this, which part am I missing here?
 
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Oh I see, they have used L'hopital rule since we got infinity over infinity
 
bigu01 said:

Homework Statement


Which of the series, diverge or converge ∑ 5^n/(4^n +3 )


Homework Equations





The Attempt at a Solution

Taking the limit as n→∞ we have (5^n ln 5)/ (4^n ln 4) , my question is here how does it become like this, which part am I missing here?

A necessary condition for \sum a_n to converge is \lim_{n \to \infty} a_n = 0 (it is not a sufficient condition; the series \sum n^{-1} diverges). Here <br /> \lim_{n \to \infty} \frac{5^n}{4^n + 3}<br /> is calculated using L'hopital's rule. Since the limit is not zero the sum does not converge.
 
If ##\Sigma a_n## converges, then ##\lim(a_n) = 0##.

If ##\lim(a_n) ≠ 0##, then ##\Sigma a_n## diverges.

Alternatively, you could apply the comparison test + the geometric test with ##|r| ≥ 1##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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