Determining if a series converges

Sunwoo Bae
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Homework Statement
Does the following series converge? Give reasons. (Series shown below)
Relevant Equations
None
323105F3-C6F9-4CE1-A68C-B548847BA194.jpeg

The following is my attempt at the solution.
Here, I used limit comparison test to arrive at the answer that the series converges.
However, the answer sheet reads that the series diverges.
I am confused because I cannot figure where my work went wrong…
can anyone tell me how the series diverges, and why my work is incorrect?

Thank you!
 
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The geometric series with terms ##(\frac 5 4)^n## clearly diverges. The formula for a convergent geometric series does not apply. And, in particular, the sum of the series in not ##-5##. That would be too absurd!
 
PeroK said:
The geometric series with terms ##(\frac 5 4)^n## clearly diverges. The formula for a convergent geometric series does not apply. And, in particular, the sum of the series in not ##-5##. That would be too absurd!
Got it! Thank you!
 
PeroK said:
That would be too absurd!
As opposed to just absurd enough? :wink:
 
Sunwoo Bae said:
Homework Statement:: Does the following series converge? Give reasons. (Series shown below)
Relevant Equations:: None

However, the answer sheet reads that the series diverges.
I am confused because I cannot figure where my work went wrong…
In your work for the Limit Comparison Test, you arrived at a limit of 1. Your mistake was not realizing that ##\sum \frac {5^n}{4^n}## is a divergent series. Since the limit you calculated was positive and finite, the series you were working with had the same behavior as ##\sum \frac {5^n}{4^n}##.

Another test that is sometimes useful is the Nth Term Test for Divergence. Since ##\lim_{n \to \infty} \frac {5^n}{4^n + 3} = \infty## (work not shown), then the series ##\sum_{n \to \infty} \frac {5^n}{4^n + 3}## diverges. Any time you get a limit that isn't 0 or fails to exist in this test, the series diverges.
 
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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