Let $a_{n+1} = a_n + \frac{1}{{a_n}^{2005}}$ and $a_1=1$. Show that $\sum^{\infty}_{n=1}{\frac{1}{n a_n}}$ converge.

Question

Let $$a_{n+1} = a_n + \frac{1}{{a_n}^{2005}}$$ and $$a_1=1$$. Show that $$\sum^{\infty}_{n=1}{\frac{1}{n a_n}}$$ converge.

Attempt

Let$$\frac{1}{na_n}=b_n$$ then because

$$a_{n+1} = a_n + \frac{1}{{a_n}^{2005}}$$

we have

$$\frac{1}{\left(n+1\right)b_{n+1}}=\frac{1}{nb_{n}}+\left(nb_n\right)^{2005}$$

Hence $$b_{n+1}=\frac{nb_n}{\left(n+1\right)\left(1+\left(nb_n\right)^{2006}\right)}$$

that means

$$\frac{b_{n+1}}{b_n}=\frac{n}{\left(n+1\right)\left(1+\left(nb_n\right)^{2006}\right)}$$

As the $$lim_{n \to \infty}\frac{b_{n+1}}{b_n}=lim_{n \to \infty} \frac{n}{n+1} lim_{n \to \infty} \frac{1}{1+\left(nb_n\right)^{2006}}=lim_{n \to \infty} \frac{1}{1+\left(nb_n\right)^{2006}}$$

Now note that $1+\left(nb_n\right)^{2006}>1$ because $\left(nb_n\right)^{2006}>0$

Hence

$$lim_{n \to \infty}\frac{b_{n+1}}{b_n}<1$$

By ratio test $\sum^{\infty}_1 b_n$ converges

Answer 1

You just have to show that for some $\varepsilon>0$ we have $a_n\gg n^\varepsilon$.
The continuous analogue of the difference equation $a_{n+1}-a_n = a_n^{-2005}$ is the differential equation $f'(x) f(x)^{2005} = 1$, with a solution given by $f(x)=(2006 x)^{\frac{1}{2006}}$. In particular $\varepsilon=\frac{1}{2006}$ looks like a good candidate. Let $A_n=a_n^{2006}$. What can you prove about $A_{n+1}-A_n$ through Bernoulli's inequality? If you manage to show that $A_{n+1}-A_n \geq c > 0$ for any $n$ large enough you are done.