/sci/ - Science & Math

>>7749140
Yes

>>7749140
No

It does converge

>>7749140
It diverges. Use the comparison test.

this converges to a fraction for sure

>>7749140
The fraction approaches 0, but the 1 is still being added in the sum, so it doesn't converge.

It does not converge

>>7749140

>>7749140

the derivative of log(k) is 1/k. Hence 1-log(k)/log(k+1)= (log(1+k)-log(k)/log(k+1)) is like 1/(k log(k+1)).

That last one diverges because the log function, to use an english cliche, goes nowhere fast.

[math] 1 - \frac { \log(k) } { \log(k+1) } = \frac { 1 } { \log(k+1) } ( \log(k+1) - \log(k) ) = \frac { 1 } { \log(k+1) } \log( 1 + \frac {1} {k} ) > \frac { 1 } { \log(k+1) } > \frac {1} {k} [/math]

[math] \sum_{k=2}^\infty ( 1 - \frac { \log(k) } { \log(k+1) } ) > \sum_{k=2}^\infty \frac {1} {k} =\infty [/math]

>>7749208
nice.

>>7749200
now that I think of it, the "is like" here can be replaced with a < so it's a proof

>>7749208
log(1+1/k) > 1
is false

>>7749208
>log(1+1/k) > 1

Not true. In fact, for every integer k log(1+1/k) < 1.

>>7749270
Oh, you're right of course

>>7749270
>>7749291
So let me try again

We want to show that the sum over
[math] \frac { 1 } { \log (1+k) } \, \log(1+1/k) [/math]
is divergent.

[math] \frac { 1 } { \log (1+k) } \, \log(1+1/k) = \frac {k} {k} \frac { 1 } { \log (1+k) } \, \log(1+1/k) = \frac { 1 } { k \, \log (1+k) } \, \log((1+1/k)^k) [/math]

The latter factor is bounded by 1 as [math] \lim_{k \to \infty} (1+1/k)^k = e^1 [/math] but always larger than the number [math] 2 \log(3/2) > 0 [/math], when k=2.

We have reduced the problem to showing that the sum over
[math] \frac { 1 } { k \, \log (1+k) } [/math]
or
[math] \frac {d} {dk} \log(\log(k)) [/math]
is divergent.

It's true by passing to the integral and using the fundamental theorem of calculus. The thing diverges only log-log and that's why it's hard.

>>7749184
>The fraction approaches 0
No.
>>7749196
Thank you.
>>7749200
Replacing the differences of log(k) by its derivative is interesting but not clear for me.
>That last one diverges because the log function, to use an english cliche, goes nowhere fast.
What do you say about the marked function?
>>7749208
I think you was joking.

[math] k > \frac {1} {k} [/math]

[math] \sum_{k=2}^\infty k > \sum_{k=2}^\infty \frac {1} {k} =\infty [/math]

>>7749354
Sadly I can't find the reference, but the general result to where the boundary between convergence and divergence lies is that

[math] \sum_{k=0}^\infty \frac {1} {k} \frac {1} { \log(k)\ \log_2 (k)\ ··· \ \log_{n-1}(k)\ (\log_n (k))^s } [/math]

is convergent for s>1.

In particular, regarding your pic,

[math] \sum_{k=0}^\infty \frac {1} {k} \frac {1} { \log(k)^s} [/math]

diverges only for s=1

(what the sum in this question comes down to)

>>7749354
Sadly I can't find the reference, but the general result to where the boundary between convergence and divergence lies is that

[math] \sum_{k=2}^\infty \frac {1} {k} \frac {1} { \log(k)\ \log_2 (k)\ ··· \ \log_{n-1}(k)\ (\log_n (k))^s } [/math]

is convergent for s>1.

In particular, regarding your pic,

[math] \sum_{k=2}^\infty \frac {1} {k} \frac {1} { \log(k)^s} [/math]

diverges only for s=1

(what the sum in this question comes down to)

>>7749350
>>7749373
Thank you man.

[eqn] \sum_{k=2}^\infty \left( 1 - \frac{\log(k)}{\log(k+1)} \right) [/eqn]
[eqn] = \sum_{k=2}^\infty \frac{1}{\log(k+1)} \int_k^{k+1} \frac{1}{x} dx [/eqn]
[eqn] = \sum_{k=2}^\infty \int_k^{k+1} \int_1^{k+1} \frac{1}{xy} dy dx [/eqn]
[eqn] \geq \int_2^\infty \int_1^{1+x} \frac{1}{xy} dy dx [/eqn]
[eqn] = \int_2^\infty \frac{\log(x+1)}{x} dx [/eqn]
[eqn] = \infty [/eqn]


It diverges.

>>7749354
>>7749373
Huh. The integral of (n ln(n))^-1 is ln(ln(n)) which goes to infinity. But the integral of n^-1 * ln(n)^(-(1+epsilon)) is -((epsilon)^-1) *log(n)^(-epsilon) which converges. I haven't wished to have more intuition with calculus formulas in a long time.

>>7749453
is that a question?

>>7749512
No. I understand it but it's unintuitive. Whenever I read a proof I try to distill it down to a central trick and doing that with a proof by derivative formulas backwards is annoying.

Now that I look at it enough there's a sweet proof of the divergence by stirling's approximation so I can be satisfied.

>>7749512
No. I understand it but it's unintuitive. Whenever I read a proof I try to distill it down to a central trick and doing that with a proof by derivative formulas backwards is annoying.

Now that I look at it enough there's a sweet proof of the divergence and convergence by stirling's approximation so I can be satisfied.

This thread prompted me to look into
[math]\displaystyle \frac{1}{k~log(k)}[/math]
and discover that Wolfram's "comparison test" is a comparison to the integral, which I've always knows as the "integral test". Am I behind the curve, or is Wolfram retarded?

>>7750264
The integral test is just a comparison test desu, see proof on wikipedia.
I could also argue that both ratio and root tests are also comparison tests tho lol

>>7750264
The Cauchy condensation test is easier for that one, in my opinion.

>>7751072
see Example 3:
http://thinkingmachineblog.net/cauchy-condensation-test/

>>7749350
Just a question, why does 1/k*log(1+k) = d/dk([log(log(k))]. When I differentiate log(log(k)) I'm getting 1/(klog(k)). Where does the +1 come from? Probably a stupid question..

>>7749140
It does not converge since the limit of the function is 1, and an infinite series with 1 as the function diverges.

>>7751439
>why does 1/k*log(1+k) = d/dk([log(log(k))]
It doesn't.
[math]\displaystyle \frac{1}{k~ \ln(1+k)}= \frac{k+1}{k} \times \frac{d}{dk} \ln \ln(1+k)[/math]

>>7752110
It's not because the limit of the function = 1 that the series does not converge.
e.g.
sum n = -1/12. Which follows from the comlex completion of sum n^-p.

But indeed, in the context of basic calculus given series diverges. I don't know if it does in complex calculus tho, and honestly I don't have the time to check it :p

>>7752415
Shut the fuck up, dumb shit

>>7749145
>>7749147
> answering twice to not get it wrong
> top kek samefag

>>7749145
>>7749147
>>7752656
>>7752423

Please no fighting. Both answers are equally correct.

>>7752110
I'm retarded. The limit of the function is 0. The series converges somewhere.

>>7753475
like
[math]\sum_{n=1}^\infty \frac {1} {n} [/math]

>>7753123
>That pic

>>7751072
Cauchy condensation test is best test

>>7753475
>The series converges somewhere
... in a galaxy far, far away, where Mathematics is somewhat different from what it is here on Earth.