# About the divergence of $\sum^\infty_{n=1}\frac 1 {n\cdot n^{1/n}}$

There's a rule of thumb that if $a>1$ then the series: $\displaystyle\sum^\infty_{n=1}\frac 1 {n^a}$ converges.

Now the series: $\displaystyle\sum^\infty_{n=1}\frac 1 {n\cdot n^{\frac 1 n}}=\sum^\infty_{n=1}\frac 1 { n^{1+\frac 1 n}}$ diverges (by the limit comparison test) but we have here a series such that the exponenet is larger than $1$ for all $n$.

So my question is why is the rule of thumb doesn't work in this case?

This is a great question! I remember thinking about this when I was working through calculus. The problem with the sum $$\sum^\infty_{n=1}\frac 1 { n^{1+\frac 1 n}}$$ is that it becomes too similar to the harmonic series for large enough $n$. The limit comparison test is a tool that allows us to see this. I will also have to admit that it does seem counter-intuitive on first glance. The result you cited about $$\sum^\infty_{n=1}\frac 1 {n^a}$$ converging for all $a>1$ relies on the fact that $a$ is constant. We can always find a $K \in \Bbb{N}$ such that $1+\frac{1}{k}<a$ for all $a>1$ and for all $k>K$, and that will ruin convergence for the series in question.

The reason is that the exponent of $n$ in your series varies with $n$. (Note that in the rule of thumb, the exponent of $n$ is constant.)

I suppose it is because in your rule of thumb $a$ is independent of $n$. More importantly,

$$\inf_n 1+\frac{1}{n} = 1$$

makes comparison test not useful in this example.

• I'm not familiar with the $inf$ over $n$ notation... – GinKin Mar 2 '15 at 17:31
• @GinKin for all intents and purposes, that is to say that $\lim_{n \to \infty} 1+\frac{1}{n} = 1$. – graydad Mar 2 '15 at 17:32
• @GinKin If the infimum (limit) of the exponent is slightly larger than 1, say $1.0001+\frac{1}{n}$, the series would have converge by comparison test and your rule of thumb. – Empiricist Mar 2 '15 at 17:38
• But by comparing it to the harmonic series, we get a a limit of $1$, thus it diverges like the harmonic series. – GinKin Mar 2 '15 at 17:40