Use the integral test to determine if $\displaystyle \sum_{n=4}^\infty \frac{n^2}{(n^3 + 4)^5}$ converges or diverges.
What integral do you need to compute to determine if this series converges or diverges?
If we replace the $n$ by $x$ in the expression and integrate, we need to know if $\displaystyle \int_4^\infty \frac{x^2}{(x^3 + 4)^5}\ dx$ converges or diverges.
Use the substutition $u = x^3 + 4$ and see how that helps.
Since $\displaystyle \int_4^\infty \frac{x^2}{(x^3 + 4)^5}\ dx$ converges, the series $\displaystyle \sum_{n=4}^\infty \frac{n^2}{(n^3 + 4)^5}$ converges.
Use the Direct Comparison Test to determine if $\displaystyle \sum_{n=3}^\infty \frac{n^2 + 3n + 2}{n^3 - 3}$ converges or diverges.
Based on the terms on the top and bottom of this fraction, should this series converge or diverge?
The gut reaction is that this series should diverge, since $n^2$ over $n^3$ should go like $1/n$. How can I make this series smaller and still diverge?
If we remove the non-dominant terms from the numerator and denominator, we can use the fact that $n^2 + 3n + 2 > n^2$ and $n^3 - 3 < n^3$ to give that $\displaystyle \frac{n^2 + 3n + 2}{n^3 - 3} > \frac{n^2}{n^3} = \frac{1}{n}$.
What do we know about the series for $\frac{1}{n}$?
Since $\displaystyle \frac{n^2 + 3n + 2}{n^3 - 3} > \frac{n^2}{n^3} = \frac{1}{n}$ and $\displaystyle \sum_{n=3}^\infty \frac{1}{n}$ diverges, we know that $\displaystyle \sum_{n=3}^\infty \frac{n^2 + 3n + 2}{n^3 - 3}$ diverges.
Use the Limit Comparison Test to determine if $\displaystyle \sum_{n=2}^\infty \frac{n^4 - 3n^2 + 10}{(n^3 + 3n^2 + 2n + 1)^2}$ converges or diverges.
Think about what the dominant terms in the numerator and denominator are. Use this to decide what $b_n$ should be.
Based on this, we should look at $\displaystyle b_n = \frac{n^4}{n^6} = \frac{1}{n^2}$, which means we will hopefully prove that this series converges.
Look at $\displaystyle L = \lim_{n\rightarrow \infty} \frac{a_n}{b_n}$ and see what this gives you.
Since $L = 1$ what does that mean for the series?
By the Limit Comparison Test with the series $\displaystyle \sum_{n=2}^\infty \frac{1}{n^2}$, we see that $\displaystyle \sum_{n=2}^\infty \frac{n^4 - 3n^2 + 10}{(n^3 + 3n^2 + 2n + 1)^2}$ converges.
Determine if the series $\displaystyle \sum_{n=1}^\infty \frac{(\ln(n))^{25}}{n^2}$ converges or diverges.
A key fact for proving this is the fact that for any positive power $a$, $n^a > \ln(n)$ for $n$ large enough. (Like the large enough that we need for the Direct Comparison Test)
You can use this fact to show that $n^{1/50} > \ln{n}$ for $n$ large, so that $n^{1/2} > (\ln{n})^{25}$ for $n$ large enough.
This fact allows us to rewrite $\displaystyle \frac{(\ln(n))^{25}}{n^2}$ as $\displaystyle \frac{1}{n^{3/2}} \frac{(\ln(n))^{25}}{n^{1/2}}$, and for $n$ large, this last factor is less than 1.
Therefore, we know that for $n$ large enough, $\displaystyle \frac{(\ln(n))^{25}}{n^2} < \frac{1}{n^{3/2}}$, and we know what happens to that series.
By the Direct Comparison Test with the series $\displaystyle \sum_{n=1}^\infty \frac{1}{n^{3/2}}$, the series $\displaystyle \sum_{n=1}^\infty \frac{(\ln(n))^{25}}{n^2}$ converges.
Determine if the series $\displaystyle \sum_{n=3}^\infty \frac{1}{n(\ln(n))^2 - n}$ converges or diverges.
First, think about the series $\displaystyle \sum_{n=3}^\infty \frac{1}{n(\ln(n))^2}$, since this should be the dominant term for the series we care about.
Use the integral test to analyze this series.
With the substitution $u = \ln(n)$, the integral converges, and so the series converges.
Now we can go back to the original series. How can we compare this series to the one we already know converges? Does the inequality go in the right direction to use direct comparison?
Using the Limit Comparison test with $\displaystyle \sum_{n=3}^\infty \frac{1}{n(\ln(n))^2}$ (with limit $L = 1$), we see that $\displaystyle \sum_{n=3}^\infty \frac{1}{n(\ln(n))^2 - n}$ converges.
Determine if the series $\displaystyle \sum_{n=6}^\infty \frac{\sin(1/n)}{\sqrt{n}}$ converges or diverges.
This is a series with positive terms because as $n \rightarrow \infty$, $\sin(1/n) \rightarrow 0$ and is positive on this range. What can we do with $\sin(1/n)$ as $n \rightarrow \infty$?
We know that $\sin(1/n)$ goes to zero, but we actually know more than that. From calculus 1, we know that $\lim_{n\rightarrow \infty} n \sin(1/n) = 1$ because $\displaystyle \lim_{x \rightarrow 0} \frac{\sin{x}}{x} = 1$.
Since we know things about limits, we should try the limit comparison test here. Since $\lim_{n\rightarrow \infty} n \sin(1/n) = 1$, we know that $\displaystyle \lim{n\rightarrow \infty} \frac{\sin(1/n)}{\sqrt{n}} = \lim_{n\rightarrow \infty} \frac{n\sin(1/n)}{n^{3/2}}$ which behaves like $\displaystyle \frac{1}{n^{3/2}}$.
Do a limit comparison test with $\frac{1}{n^{3/2}}$ to see what you get. The results in the previous hint tells you the limit here is 1.
By limit comparison test with the series $\displaystyle b_n = \frac{1}{n^{3/2}}$, we see that the series $\displaystyle \sum_{n=6}^\infty \frac{\sin(1/n)}{\sqrt{n}}$ converges.
Determine if the series $\displaystyle \sum_{n=4}^\infty \frac{1}{n\ln{n}}$ converges or diverges.
Determine if the series $\displaystyle \sum_{n=1}^\infty \frac{\cos^2{n}}{n^2}$ converges or diverges.