If I use the table mode on my Casio and choose a very large step and end ( in excess of over 30 million terms) it doesn't seem like it wants to converge.
[Edit]
Oops, I forgot you wanted to sum over the variable.
It gets bigger without limit.
Does converge?
Root test got a 1, I wasn't able to do the ratio test (algebraic ugliness never cleared up), we can't integrate it (right?), I don't know what to compare it to, so the freshman calculus stuff isn't working.
I thought about factoring out a piece like and applying Dirichlet's test (but the factored out part doesn't go to zero) or Abel's test (but the remaining ^(n-1) part may or may not converge.)
I thought about (abstractly) converting this to a power series, calling it equal to the original sum, and taking derivatives to see if I could find something easier to identify a radius of convergence, but with the power series I think I get into conditional convergence and the effect of rearranging terms..
Hm..
Any ideas?
Thank you!
[EDIT: You may skip some inconclusive dialog and get to my latest attempt to prove divergence here at post #11.]
If I use the table mode on my Casio and choose a very large step and end ( in excess of over 30 million terms) it doesn't seem like it wants to converge.
[Edit]
Oops, I forgot you wanted to sum over the variable.
It gets bigger without limit.
Perhaps instead of "does this converge" I should say, "show that this diverges or converges".
Wolfram|Alpha says it doesn't converge by comparison test, but of course it doesn't tell what to compare it to:
sum(1 - (log n)/n)^n
Because if I consider on it's own.
Then at 2 trillion terms it is
For two trillion terms the decimal point has only shifted 6 places. The sum just continues to grow.
In addition, if I consider the expression for the term (s) alone from the sum you just gave me, that has gone to zero way before it has even reached two trillion terms.
I appreciate the effort you are making, but the things you are saying are just not correct. For both series given, the sums "continue to grow", but the limit of one is a real number while the other has no limit. It seems you are trying to do analysis by numerical methods, perhaps on your calculator.
Maybe someone of greater understanding will reply to this thread and better explain the problem with your technique, and perhaps, I hope, actually answer my question.
Hmm.. How about this..
Take as a given that is an increasing sequence with . (It is true, I'm just not proving it here.)
For we know that
but that same sum diverges for any other r.
Let . Then, by definition of limit, there exists such that for each , . Thus for any , there are an infinite number of terms such that
. Hence, is greater than any (monic) convergent geometric series, and is therefore divergent.
QED?
I completely understand understand why you say that
Because it is completely false. Look at this.