We will use the Comparison Test to prove this result. We must note the following:

\(\displaystyle \frac{n}{n^2+1}\)  is positive. 

We have all natural numbers n:

\(\displaystyle n^2\ge n\) , this implies that

\(\displaystyle n^2+1\ge n+1 > n\).

Inverting we get :

\(\displaystyle \frac{1}{n} < \frac{n}{n^2+1}\)

Summing from 1 to \(\displaystyle \infty\), we have

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n}{n^2+1}\)

We know that the \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\) is divergent. Therefore by the Comparison Test:

\(\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^2+1}\) is divergent.

We will use the comparison test to prove that

\(\displaystyle \small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}\)

converges (Note: we cannot use the ratio test, because then the ratio will be \(\displaystyle \small 1\), which means the test is inconclusive).

We will compare \(\displaystyle \small \frac{n^3}{n^7+1000}\) to \(\displaystyle \small \frac{1}{n^4}\) because they "behave" somewhat similarly. Both series are nonzero for all \(\displaystyle \small n\geq 1\), so one of the conditions is satisfied.

The series 

\(\displaystyle \small \small \sum_{n=1}^\infty \frac{1}{n^4}\)

converges, so we must show that 

\(\displaystyle \small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}\) 

for \(\displaystyle \small n\geq 1\).

This is easy to show because

\(\displaystyle \small \frac{1}{n^4}=\frac{n^3}{n^7}\geq \frac{n^3}{n^7+1000}\)

since the denominator \(\displaystyle \small n^7+1000\) is greater than or equal to \(\displaystyle \small n^7\) for all \(\displaystyle \small n\geq 1\).

Thus, since 

\(\displaystyle \small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}\)

and because

\(\displaystyle \small \small \sum_{n=1}^\infty \frac{1}{n^4}\)

converges, it follows that 

\(\displaystyle \small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}\)

converges, by comparison test.

We can compare this to the series \(\displaystyle \sum_{n = 1}^{\infty } \frac{1}{n^{3}}\) which we know converges by the p-series test.

To figure this out, let's first compare \(\displaystyle n^{3} + 2\) to \(\displaystyle n^{3}\). For any number n, \(\displaystyle n^{3} + 2\) will be larger than \(\displaystyle n^{3}\).

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, \(\displaystyle n^{3} + 2 > n^{3}}\) turns into 

\(\displaystyle \frac{1}{n^{3}} > \frac{1}{n^{3} + 2}\).

And so, because \(\displaystyle \frac{1}{n ^{3}}\) converges, thus our series also converges. 

We can solve this problem quite simply with the integral test. We know that if 

\(\displaystyle \int_1^{\infty} \frac{1}{x^p} dx\)

converges, then our series converges. 

We can rewrite the integral as 

\(\displaystyle \int x^{-p} dx\)

and then use our formula for the antiderivative of power functions to get that the integral equals

\(\displaystyle \frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty\).

We know that this only goes to zero if \(\displaystyle 1-p < 0\). Subtracting p from both sides, we get

\(\displaystyle 1 < p\).

We compare this series to the series \(\displaystyle \sum_{n=3}^{\infty}\frac{1}{n^2}\)

Because 

\(\displaystyle n^2+2>n^2\)  for  \(\displaystyle n=3,4,5,...\) 

it follows that

\(\displaystyle \frac{1}{n^2+2}< \frac{1}{n^2}\)  for  \(\displaystyle n=3,4,5,...\) 

This implies

\(\displaystyle \sum_{n=3}^{\infty}\frac{1}{n^2+2}< \sum_{n=3}^{\infty}\frac{1}{n^2}\)

Because the series on the right has an exponent \(\displaystyle 2>1\),

the series on the right converges

\(\displaystyle \sum_{n=3}^{\infty}\frac{1}{n^2+2}< \sum_{n=3}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{5}{4}\)

making

\(\displaystyle \sum_{n=3}^{\infty}\frac{1}{n^2+2}\) 

converge as well.

We compare this series to the series \(\displaystyle \sum_{n=2}^{\infty}\frac{1}{n}\)

Because 

\(\displaystyle ln\,n< n\)  for  \(\displaystyle n=2,3,4,...\) 

it follows that

\(\displaystyle \frac{1}{ln\,n}>\frac{1}{n}\)  for  \(\displaystyle n=2,3,4,...\) 

This implies

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

Because the series on the right has a degree of \(\displaystyle n\) equal to \(\displaystyle 1\) in the denominator,

the series on the right diverges

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

making

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

diverge as well.

To show this series converges, we use direct comparison with

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

which converges by the p-series test with \(\displaystyle p=2>1\).

Thus we must show that

\(\displaystyle \frac{1}{n(n+1)}\le\frac{1}{n^2}\).

Cross multiplying the previous section and multiplying \(\displaystyle n\) by \(\displaystyle n+1\), we obtain \(\displaystyle n^2\le{n^2+n}\leftrightarrow{n\ge0}\).

Since this holds for all \(\displaystyle n=1,2,3,...\) we can conclude that 

\(\displaystyle \frac{1}{n(n+1)}\le\frac{1}{n^2}\).

Summing from \(\displaystyle 1\) to \(\displaystyle \infty\), and noting that 

\(\displaystyle \frac{1}{n(n+1)}>0\) for all \(\displaystyle n=1,2,3,...\), we obtain the following inequality:

\(\displaystyle 0< \sum_{n=1}^\infty\frac{1}{n(n+1)}< \sum_{n=1}^\infty\frac{1}{n^2}\).

Therefore the series 

\(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)}\)

converges by direct comparison.

Now to find the value, we note that 

\(\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\),

so that

\(\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)}=\sum_{n=1}^\infty\biggr(\frac{1}{n}-\frac{1}{n+1}\biggr)\).

Now let 

\(\displaystyle s_k=\sum_{n=1}^k\biggr(\frac{1}{n}-\frac{1}{n+1}\biggr)\)

be a sequence of partial sums.

Then we have \(\displaystyle s_k=\sum_{n=1}^k\biggr(\frac{1}{n}-\frac{1}{n+1}\biggr)\)

\(\displaystyle =\biggr(1-\frac{1}{2}\biggr)+\biggr(\frac{1}{2}-\frac{1}{3}\biggr)+\biggr(\frac{1}{3}-\frac{1}{4}\biggr)+...+\biggr(\frac{1}{k-1}-\frac{1}{k}\biggr)\)

\(\displaystyle =1-\frac{1}{k}\)

Therefore

\(\displaystyle s_k=1-\frac{1}{k}\).

Taking the limit as \(\displaystyle k\rightarrow\infty\), we obtain the following:

\(\displaystyle \lim_{k\rightarrow\infty}s_k=\lim_{k\rightarrow\infty}\sum_{n=0}^k\frac{1}{n(n+1)}=\sum_{n=0}^\infty\frac{1}{n(n+1)}=\lim_{k\rightarrow\infty}\biggr(1-\frac{1}{k}\biggr)=1-\frac{1}{\infty}=1\)

Therefore we have

\(\displaystyle \sum_{n=0}^\infty\frac{1}{n(n+1)}=1\).

Notice that \(\displaystyle n-1< n\) for \(\displaystyle n=2,3,...\)

This implies that

\(\displaystyle \frac{1}{n-1}>\frac{1}{n}\) for \(\displaystyle n=2,3,...\)

Which then  implies

\(\displaystyle \sum_{n=2}^{\infty}\frac{1}{n-1}>\sum_{n=2}^{\infty}\frac{1}{n}\)

Since the right-hand side is the harmonic series, we have

\(\displaystyle \sum_{n=2}^{\infty}\frac{1}{n-1}> \infty\)

and thus the series does NOT converge.