Find a sequence $a_n \downarrow 0$ such that lim$_{n\to\infty}na_n=0$ but such that $\sum_{n=1}^{\infty}a_n$ diverges. My professor gave $a_n=\frac{1}{n\log n}$ for $n \geq 2$ as such an example but I can't understand why this is true. I am more accepting that the series $\sum_{n=2}^{\infty}\frac{1}{n\log n}$ is actually convergent rather than divergent. Here is my reasoning: 
Note that $\sum_{n=2}^{10}\frac{1}{n\log n}$ is some finite number. 
Note that $\sum_{n=2}^\infty \frac{1}{n\log n} = \sum_{n=2}^{10}\frac{1}{n\log n} + \sum_{n=11}^{\infty}\frac{1}{n\log n}$. 
But, $n\log n > n$ for $n>10$. I can let $n\log n =n^p$ where $p>1$. Then, using the p-series, $\sum_{n=11}^{\infty}\frac{1}{n\log n} = \sum_{n=11}^\infty \frac{1}{n^p}$ converges to some finite number. 
Therefore, $\sum_{n=2}^\infty \frac{1}{n\log n} = \sum_{n=2}^{10}\frac{1}{n\log n} + \sum_{n=11}^\infty \frac{1}{n\log n}$ must converge. $\blacksquare$
My professor said this example was to give us an intuition that $p$-series will work for any $p>1$ even though $p=1.0000000000000001$, or $p$ is greater than $1$ by very very small amount. 
 A: For a fixed value of $n,$ you can chose $p>1$ so that $n\log n = n^p,$ but for the next value of $n$ you get a different value of $p$, and for the next you get yet another value of $p,$ and so on. The value of $p$ does not stay fixed as $n$ changes. That is the error in your argument.
An integral test handles this quickly:
$$
\int \frac{dx}{x\log x} = \int \frac{du} u \text{ where } u = \log x \text{ and } du = \frac{dx} x.
$$
A: If $n\log(n)=n^p$ for $n>10$ then $\log(n)=n^{p-1}$ for $n>10$, which is false. $\log(n)$ grows more slowly than any positive power of $n$.
To prove that $\sum\frac{1}{n\log n}$ diverges, you can use the integral test or Cauchy's condensation test.
A: Suppose that we agree $\ln(n)<n$.  It then follows that for any $p>0$,
$$\ln(n^p)<n^p$$
But by log rules, this means
$$\ln(n)<\frac1pn^p$$
or,
$$\ln(n)<<n^p$$
For every $p>0$.  I mean, sure, it's true that $n<n\ln(n)<n^{1+p}$ for every $p>0$, but this does not mean the same thing as $n\ln(n)=n^p$.
Taking it along your lines, I could say that it grows slower than any $n^{1+p}$, thus it diverges by the p-series, but as we will find, this logic is still wrong.

To me, I find the Cauchy condensation test very intuitive:
$$\begin{align}S&=\frac1{2\log(2)}+\frac1{3\log(3)}+\frac1{4\log(4)}+\frac1{5\log(5)}+\dots\\&>\frac1{2\log(2)}+\frac1{4\log(4)}+\frac1{4\log(4)}+\frac1{8\log(8)}+\dots\end{align}$$
We replace every term with $2^n$ such that all terms will be smaller than or equal to the original.  After doing that, we have one $2$, two $4$'s, four $8$'s, ..., $2^{n-1}$ amount of $2^n$'s.  We can then simplify this into the next line:
$$\sum_{n=2}^\infty\frac1{n\log(n)}>\sum_{n=1}^\infty\frac{2^{n-1}}{(2^n)\log(2^n)}=\frac1{2\log(2)}\sum_{n=1}^\infty\frac1n$$
Now, it's not so hard to see that the series diverges by the p-series.

Lesson here?  It diverges logarithmically slower than $\sum_{n=1}^\infty\frac1n$, which is why divergence was not obvious.  Oddly, diverging "logarithmically slower" hides any obvious comparison to the p-series, as it lies somewhere between $\frac1n$ and $\frac1{n^{1+p}}$.

As a side note, a table of series involving logarithms and such:
$$\begin{array}{c|c}\text{diverges}&\text{converges}\\\hline\frac1{n\log(n)}&\frac1{n\log^2(n)}\\\frac1{n\log(\log(n))}&\frac{(-1)^n}{\log(n)}\\\frac1{\log(n!)}&\frac1{n\log^{1.1}(\log(n))}\end{array}$$
