Use the Ratio Test to determine if the series $\displaystyle \sum_{n=2}^\infty \frac{n^{40}}{n!}$ converges or diverges.
Take the ratio of consecutive terms of this series. How does this ratio simplify?
The limit we need to take for the ratio test is $\displaystyle \lim_{n\rightarrow \infty} \frac{(n+1)^{40}}{n^{40}}\frac{1}{n+1}$.
Since this limit is zero, what does this say about the series?
This series converges by the Ratio test.
Use the Root Test to determine if the series $\displaystyle \sum_{n=3}^\infty \left(2 + \frac{1}{n}\right)^{-n}$ converges or diverges.
We need to take the nth root of the terms of the series, and then take the limit.
If we do that, the limit $t$ we are looking for is $\displaystyle \lim_{n \rightarrow \infty} \left(2 + \frac{1}{n}\right)^{-1}$.
With this, we have that $t = \frac{1}{2}$. What does that say about the series?
This series converges by the Root test.
Determine whether the series $\displaystyle \sum_{n=4}^\infty (0.8)^{-n} n^{-0.8}$ converges or diverges by any method covered so far.
We see numbers raised to the nth power. Which test should be applied here?
We should apply the ratio test.
The limit here you need to compute for this test is $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{0.8} \frac{n^{0.8}}{(n+1)^{0.8}}$ What happens as $n \rightarrow \infty$?
The limit goes to $\frac{5}{4}$. What does this mean for the series?
This series diverges by the ratio test.
Determine whether the series $\displaystyle \sum_{n=3}^\infty (-1)^n \cos\left(\frac{1}{n}\right)$ converges or diverges by any method covered so far.
The $(-1)^n$ suggests that this is an alternating series. Does the Alternating Series test apply?
For this, the sequence $b_n$ would be $\cos\left(\frac{1}{n}\right)$. Is this sequence decreasing? Does it go to zero?
This sequence does not go to zero. So the alternating series test does not apply. What else can we do?
We already saw that the terms don't go to zero. What does that tell us?
This series diverges by the nth term divergence test.
Determine whether the series $\displaystyle \sum_{n=3}^\infty \frac{n^2 + 5n}{7n^4 + 9}$ converges or diverges by any method covered so far.
Look at the dominant terms on the top and bottom of the terms of this series. What does this suggest?
This suggests that the series converges and that we could prove it by the limit comparison test.
Do the limit comparison test with the series $\displaystyle b_n = \frac{1}{n^2}$.
The limit $L = \frac{1}{7}$, so what does that mean about the series?
This series converges by the limit comparison test with $b_n = \frac{1}{n^2}$.
Determine whether the series $\displaystyle \sum_{n=5}^\infty \frac{1}{n (\ln(n))^3}$ converges or diverges by any method covered so far.
The presence of a logarithm indicates that integral test may be the way to go here, since other comparison tests will not work.
The integral needed is $\displaystyle \int \frac{1}{x (\ln(x))^3}\ dx$. Try the substitution $u = \ln(x)$.
After this substitution, the integral is $\displaystyle \int_5^\infty \frac{1}{u^3}\ du$. Does this converge or diverge?
This series converges by the integral test.
Determine whether the series $\displaystyle \sum_{n=2}^\infty \frac{(n!)^3}{(3n)!}$ converges or diverges by any method covered so far.
Determine whether the series $\displaystyle \sum_{n=5}^\infty \frac{\sin{n}}{n^2}$ converges or diverges by any method covered so far.