Originally Posted by

**Mathstud28** Here is a neat problem I did yesterday.

Show that

$\displaystyle \sum_{k=2}^{n}\frac{1}{\log_k(x)}=\frac{1}{\log_{n !}(x)}$

trivial

(Pre-calc students cover your ears)

Then use it to show that

$\displaystyle \sum_{k=2}^{\infty}\frac{1}{\log_k(x)}$ converges $\displaystyle \forall{x}>1$

Does not converge for any $\displaystyle x \ne 1$, since:

$\displaystyle \sum_{k=2}^{\infty}\frac{1}{\log_k(x)}=\sum_{k=2}^ {\infty} \frac{\ln(k)}{\ln(x)}$$\displaystyle =\frac{1}{\ln(x)}\sum_{k=2}^{\infty} \ln(k)$

Some numerical evidence:

Code:

>k=2:100;
>x=2;
>s=1/logbase(x,k);
>
>S100=sum(s)
524.765
>
>
>k=2:1000;
>x=2;
>s=1/logbase(x,k);
>
>S1000=sum(s)
8529.4
>
>
>
>k=2:10000;
>x=2;
>s=1/logbase(x,k);
>
>S10000=sum(s)
118458
>

RonL