Determining if a series converges

[Sunwoo Bae]

Homework Statement

Does the following series converge? Give reasons. (Series shown below)

Relevant Equations

None

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!

 

[PeroK]

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!

 

[Sunwoo Bae]

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!

 

[vela]

That would be too absurd!

As opposed to just absurd enough?

 

[Mark44]

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.