Question

find the sum

Answer 1

\[\huge\sum_{k=2}^n\frac{1}{\log_ku}\]

Answer 2

What's u?

Answer 3

Try induction using the change of base formula to convert everything into one base

Answer 4

\[\log_{2}(u) = \frac{ \ln (u) }{ \ln (2) } \]

Answer 5

so \[\frac{ 1 }{ \log_{k}(u) } = \frac{ 1 }{ \frac{ \ln (u) }{ \ln (k) } } = \frac{ \ln (k) }{ \ln (u) }\]

Answer 6

Which implies that
\[\sum_{k=2}^{n}\frac{ 1 }{ \log_{k}(u) } = \sum_{k=2}^{n}\frac{ \ln (k) }{ \ln (u) }=\frac{ 1 }{ \ln (u) } \sum_{k=2}^{n}\ln (k)\]

Answer 7

...this feels like it's summing to +infinity. Thoughts?

Answer 8

Unless u=0.

Answer 9

Or if u=1 or u < 0 ... assuming we're dealing with real valued solutions@_@

Answer 10

So let's assume u > 1 for simplicity. What happens then?

Answer 11

The sum is unbounded?

Answer 12

It looks like it but for k > 2 we have, for fixed u > 1,
\[\log_{k+1} (u) < \log_{k}(u) \]
i.e. it's a monotonically decreasing sequence. Hmmm, but is it decreasing "fast enough?"

Answer 13

wait....
\[\log_{k+1}(u) < \log_{k} (u) \rightarrow \frac{ 1 }{ \log_{k+1}(u) } > \frac{ 1 }{ \log_{k}(u) }\]
and when k > u the inequality on the right becomes > 1

Answer 14

@__@ confusing myself now lol gonna ponder this one a little longer a bit later

Answer 15

it equals \(\large \log_u n!\) right ?

Answer 16

@domu Thanks again :)

Answer 17

@skullpatrol lol no problem. I hope I confused people only a little bit haha

Answer 18

@ganeshie8 how did you get that? I'm curious :)

Answer 19

ive just expanded the sum

Answer 20

\(\huge\sum_{k=2}^n\frac{1}{\log_ku}\)

\(\large \frac{1}{\log_2 u} + \frac{1}{\log_3 u} + \frac{1}{\log_4 u} + \frac{1}{\log_5 u} + ... \frac{1}{\log_n u}\)

\(\large \log_u 2 + \log_u 3 +\log_u 4 + \log_u 5 +... + \log_u n\)

\(\large \log_u n!\)

Answer 21

AHHH I see it : D. Thank ya @ganeshie8

Answer 22

lol perfect timing!

Answer 23

:)