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.

Step 1: Recall the convergence rule of the power series:

According to the convergence rule of the power series....

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p}\) converges as long as \(\displaystyle p>1\)

 Step 2: Compare the exponent:

Since \(\displaystyle p=3\), it is greater than \(\displaystyle 1\). Hence the series converges.

Step 3: Conclusion of the convergence rule

Now, notice that the series isn't an alternating series, so it doesn't matter whether we check for absolute or conditional convergence.

Step 1: Try and look for another function that is similar to the original function:

\(\displaystyle \frac{n^2}{1+n^3}\) looks like \(\displaystyle \frac{n^2}{n^3}=\frac{1}{n}\)

Step 2: We will now Use the Limit Comparison test

\(\displaystyle \lim_{n\to\infty}\frac{1}{n}\times\frac{(1+n^3)}{n^2}=\lim_{n\to\infty}\frac{1+n^3}{n^3}=1\)

Since the limit calculated, is not equal to 0, the given series converges by limit comparison test

The best way to answer this question would be by comparing the series to another series,\(\displaystyle b_{n}\), that greatly resembles the behavior of the original series, \(\displaystyle a_{n}\). The behavior is determined by the terms of the numerator and the denominator that approach infinity at the quickest rate. In this case:

\(\displaystyle b_{n}=\sum_{n=1}^{\infty }\frac{n^2}{n^4}=\sum_{n=1}^{\infty}\frac{1}{n^2}\)

When this series is simplifies, it simplifies to a series that converges because of the p-test where \(\displaystyle p=2>1\).

With two series and the confirmed convergence of one of those series, the limit comparison test can be applied to test for the convergence or divergence of the original series. The limit comparison test states that two series will converge or diverge together if:

\(\displaystyle \lim_{n \to \infty}\frac{a_{n}}{b_{n}}=L>0\)

Specifically:

 \(\displaystyle \lim_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim_{n \to \infty}\frac{n^2+3}{n^4+6}\cdot \frac{n^4}{n^2}=\lim_{n \to \infty}\frac{n^6+3n^4}{n^6+6n^2}\)

This limit equals one because of the fact that:

\(\displaystyle \lim_{x \to\infty }\frac{ax^2+bx}{cx^2}=\frac{a}{c}\)  if the coefficients come from the same power. 

Because the limit is larger than zero, \(\displaystyle a_{n}\) and \(\displaystyle b_{n}\) will converge or diverge together. Since it was already established that \(\displaystyle b_{n}\) converges, the original seies, \(\displaystyle a_{n}\), converges by the limit comparison test.

\(\displaystyle 0\leq\frac{1}{n^23^n}\leq\frac{1}{3^n}\) for all \(\displaystyle n\);    \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{3^n}\) is a sum of geometric sequence with base 1/3.

Therefore, said sum converges.

Then, by comparison test,  \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^23^n}\)  also converges.

Since \(\displaystyle \frac{1}{n^2+n}< \frac{1}{n^2}\) for all n>0, and \(\displaystyle \frac{1}{n^2}\) converges as a P-Series, we may conclude that \(\displaystyle \frac{1}{n^2+n}\) must also converge by the comparison test.