1. ## An easy problem

Here is a neat problem I did yesterday.

Show that

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

Then use it to show that

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

Not too hard if you see it, but neat I think .

2. 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

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)}$$\displaystyle =\frac{1}{\ln(x)}\sum_{k=2}^{\infty} \ln(k) Some numerical evidence: Code: >k=2:100; >x=2; >s=1/logbase(x,k); > >S100=sum(s) 524.765 > > >k=2:1000; >x=2; >s=1/logbase(x,k); > >S1000=sum(s) 8529.4 > > > >k=2:10000; >x=2; >s=1/logbase(x,k); > >S10000=sum(s) 118458 > RonL 3. Originally Posted by CaptainBlack trivial Does not converge for any \displaystyle x \ne 1, since: \displaystyle \sum_{k=2}^{\infty}\frac{1}{\log_k(x)}=\sum_{k=2}^ {\infty} \frac{\ln(k)}{\ln(x)}$$\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
It's not trivial for Pre-calc Students .

And yes, it was a trick question

I just said that since it equatls $\displaystyle \log_x(n!)$

Then $\displaystyle \text{As }n\to\infty\Rightarrow{\log_{x}(n!)\to\infty}$

4. Originally Posted by Mathstud28
It's not trivial for Pre-calc Students .
It is trivial in the sense that it requires routine application of what they should know about change of base for logs and how to add fractions.

And yes, it was a trick question

I just said that since it equatls $\displaystyle \log_x(n!)$

Then $\displaystyle \text{As }n\to\infty\Rightarrow{\log_{x}(n!)\to\infty}$
You said use the identity to show that the given series converges, well it does not converge, and it obviously does not without recourse to the identity.

RonL

5. Originally Posted by CaptainBlack
It is trivial in the sense that it requires routine application of what they should know about change of base for logs and how to add fractions.

You said use the identity to show that the given series converges, well it does not converge, and it obviously does not without recourse to the identity.

RonL
Apparently you never gave your students "set up questions"?