We will use the integral test to prove this result.

We need to note the following:

$\displaystyle \frac{1}{n^2+a^2}$    is positive, decreasing and $\displaystyle lim_{n\rightarrow \infty}{\frac{1}{n^2+a^2}}=0$.

By the integral test, we know that the series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}$ is the integral $\displaystyle \int_{1}^{\infty}\frac{1}{x^2+a^2}dx$.

We know that the above intgral is finite.

This means that the series

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+a^2}$ is convergent.

We know that:

$\displaystyle chn=\frac{e^n+e^{-n}}{2}$ and therefore we deduce :

$\displaystyle ch2n=\frac{e^{2n }+e^{-2n}}{2}$

We will use the Comparison Test with this problem. To do this we will look at the function in general form $\displaystyle v_{n}\equiv \frac{e^n} {e^{2n}}$. 

We can do this since,

$\displaystyle e^{-n}=\frac{1}{e^n}$ and  $\displaystyle e^{-2n}=\frac{1}{e^{2n}}$ approach zero as n approaches infinity. The limit of our function becomes,

$\displaystyle \lim_{n\rightarrow \infty}\frac{e^n+e^{-n}}{e^{2n}+e^{-2n}}=\frac{e^n}{e^{2n}}$

This last part gives us $\displaystyle v_{n}\equiv {e^{-n}}$.

Now we know that $\displaystyle \frac{1}{e}< 1$ and noting that $\displaystyle e^{-n}}$ is a geometric series that is convergent.

We deduce by the Comparison Test that the series

having general term $\displaystyle v_{n}$ is convergent.

We will use the comparison test to prove this result. We must note the following:

$\displaystyle \frac{n^3+1}{n^4}$  is positive.

We have all natural numbers n:

$\displaystyle n^4\ge n^3$ , this implies that

$\displaystyle n^4\ge n^3 \ge n$.

Inverting we get :

$\displaystyle \frac{1}{n} < \frac{n^3+1}{n^4}$

Summing from 1 to $\displaystyle \infty$, we have

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n^3+1}{n^4}$

We know that the $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the comparison test:

$\displaystyle \sum_{n=1}^{\infty}\frac{n^3}{n^4+1}$

is divergent

We will use the Limit Comparison Test to establish this result.

We need to note that the following limit

$\displaystyle \frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\displaystyle \frac{1}{n^{3/2}}$.

We know that this series is convergent because it is a p-series. (Remember that

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

To have a series that is convergent we must have that the general term of the series goes to 0 as n goes to $\displaystyle \infty$.

We have the general term:

$\displaystyle \frac{n^7}{n^7+5}$

therefore, we have $\displaystyle \lim_{n\rightarrow \infty } \frac{n^7}{n^7+5}=1$.

This means that the general term does not go to 0 .

Therefore the series is divergent.

We note first that the general term of the series is positive.

It is also decreasing and tends to 0 as n tends to $\displaystyle \infty$. We will use the Integral Comparison Test to show this result.

Note the nature of the series is the same for the integral:

$\displaystyle \int_{1}^{\infty}\frac{1}{7x+99}dx$

This last intgral is divergent because it does not equal zero.

Therefore our series is divergent as well.

We note first that we can write the general term as: 

$\displaystyle 49\frac{2^n}{7^{n}}}$

and simplifying this term one more time, we have:

$\displaystyle 49\left(\frac{2}{7}\right)^n$

We note that since $\displaystyle \frac{2}{7}< 1$,  this series is a geometric one which is convergent.

This is what we need to show here.

We know that if a series is convergent, then its general term must go to 0 as  $\displaystyle n\mapsto \infty$.

We have $\displaystyle arctan\left(\frac{n^3}{n^2+1}\right)$ is our general term in this case.

We have $\displaystyle \lim_{n\rightarrow\infty}{arctan\left(\frac{n^3}{n^2+1}\right)}=\frac{\pi}{2}$.

Since the general term does not go to 0, the series is divergent.

We will use the Limit Comparison Test to study the nature of the series.

We note first that $\displaystyle n\ge 2$, the series is positive.

We will compare the general term to $\displaystyle \frac{1}{n^8}$. 

We note that by letting $\displaystyle v_{n}=\frac{n+11}{n^9-7}$ and $\displaystyle u_{n}=\frac{1}{n^8}$, we have:

$\displaystyle \lim_{n\rightarrow \infty}\frac{u_{n}}{v_n}}=1$.

Therefore the two series have the same nature, (they either converge or diverge at the same time). 

We will use the Integral Test to deduce that the series having the general term:

$\displaystyle u_{n}=\frac{1}{n^8}$ is convergent.

Note that we know that $\displaystyle \int_{1}^{\infty}\frac{1}{x^p}dx$is convergent if p>1 and in our case p=8 .

This shows that the series having general term $\displaystyle u_{n}$ is convergent.

By the Limit Test, the series having general term $\displaystyle v_n$ is convergent.

This shows that our series is convergent.

We will use the Comparison Test to prove this result.

We need to note first that $\displaystyle \frac{1}{\ln(n)}> 0$ for $\displaystyle n\ge 2$.

We know that $\displaystyle \ln(n)< n$ , where $\displaystyle n\ge 2$.

Inverting the above inequality, we have:

$\displaystyle \frac{1}{n}< \frac{1}{\ln(n)}$.

Now we will use the Comparison Test.

We know that the series $\displaystyle \sum \frac{1}{n}$ is divergent.

Therefore,

$\displaystyle \sum_{n=2}^{\infty}\frac{1}{\ln(n)}$

is also divergent.