Does this series converge (1 - (log n)/n)^n

**buckeye1973** · December 5th 2012, 11:07 AM

Does $\sum\left(1 - \frac{\log n}{n}\right)^n$ 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 $\left(1 - \frac{\log n}{n}\right)\left(1 - \frac{\log n}{n}\right)^{n-1}$ 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.]

**astartleddeer** · December 5th 2012, 12:54 PM

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.

**buckeye1973** · December 5th 2012, 01:02 PM

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

**astartleddeer** · December 5th 2012, 01:09 PM

If you want to "show it" just sub for n. Sum two terms, sum three terms, sum four terms and so on.

**buckeye1973** · December 5th 2012, 01:15 PM

Originally Posted by astartleddeer

If you want to "show it" just sub for n. Sum two terms, sum three terms, sum four terms and so on.

The terms converge to zero, so it is perfectly reasonable to think that it might converge (aside from knowing that it doesn't). Listing off the first 30 million partial sums doesn't really show anything in any kind of rigorous manner.

**astartleddeer** · December 5th 2012, 01:37 PM

$1 +\left(1 - \frac{log(2)}{2}\right)^2 \approx 1.721$

$1 +\left(1 - \frac{log(2)}{2}\right)^2 +\left(1 -\frac{log(3)}{3}\right)^3 \approx 2.316$

$1 +\left(1 - \frac{log(2)}{2}\right)^2 +\left(1 - \frac{log(3)}{3}\right)^3 +\left(1 -\frac{log(4)}{4}\right)^4 \approx 2.837$

$1 +\left(1 -\frac{log(2)}{2}\right)^2 +\left(1 -\frac{log(3)}{3}\right)^3 +\left(1 -\frac{log(4)}{4}\right)^4 +$

$\left(1 -\frac{log(5)}{5}\right)^5 \approx 3.308$

**astartleddeer** · December 5th 2012, 01:39 PM

I forgot you wanted to sum over a variable the first time round, which is why I made an edit to my first post.

The sum diverges.

**buckeye1973** · December 5th 2012, 01:49 PM

Originally Posted by astartleddeer

$1 +\left(1 - \frac{log(2)}{2}\right)^2 \approx 1.721$
(etc.)

Again, I don't see how listing off the first few partial sums helps.

I could list off the first few partial sums of this:
$\sum(50/n^2)^n$

but that is a convergent series... so...

**astartleddeer** · December 5th 2012, 02:11 PM

Because if I consider $\left(1- \frac{log(n)}{n}\right)^n$ on it's own.

Then at 2 trillion terms it is $\approx 1.674 \times 10^-^6$

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.

**buckeye1973** · December 5th 2012, 03:29 PM

Originally Posted by astartleddeer

Because if I consider $\left(1- \frac{log(n)}{n}\right)^n$ on it's own.

Then at 2 trillion terms it is $\approx 1.674 \times 10^-^6$

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.

**buckeye1973** · December 5th 2012, 04:03 PM

Hmm.. How about this..

Take as a given that $\left\{1 - \frac{\log n}{n}\right\}$ is an increasing sequence with $\lim_{n \to \infty} \left(1 - \frac{\log n}{n}\right) = 1$ . (It is true, I'm just not proving it here.)

For $|r| < 1$ we know that

$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$

but that same sum diverges for any other r.

Let $r \in (0,1)$ . Then, by definition of limit, there exists $N \in \mathbb{N}$ such that for each $n \geq N$ , $\left(1 - \frac{\log n}{n}\right) > r$ . Thus for any $r \in (0,1)$ , there are an infinite number of terms such that
$\left(1 - \frac{\log k}{k}\right)^k > r^k$ . Hence, $\sum\left(1 - \frac{\log n}{n}\right)^n$ is greater than any (monic) convergent geometric series, and is therefore divergent.

QED?

**Plato** · December 5th 2012, 04:06 PM

Originally Posted by buckeye1973

Maybe someone of greater understanding will reply to this thread and better explain the problem

It is a simple comparison.

$\sum\limits_{n = 2}^\infty {\left( {1 - \frac{{\log (n)}}{n}} \right)^n = } \sum\limits_{n = 2}^\infty {\left( {n- \log(n)}\right)^n\left( {\frac{1}{n}} \right)^n > \sum\limits_{n = 2}^\infty {\left( 1 \right)} }$

**buckeye1973** · December 5th 2012, 04:11 PM

Thank you, Plato, but I'm not understanding why this inequality is true:

Originally Posted by Plato

$\sum\limits_{n = 2}^\infty {\left( {n- \log(n)}\right)^n\left( {\frac{1}{n}} \right)^n > \sum\limits_{n = 2}^\infty {\left( 1 \right)} }$

?

**Plato** · December 5th 2012, 04:23 PM

Originally Posted by buckeye1973

Take as a given that $\left\{1 - \frac{\log n}{n}\right\}$ is an increasing sequence with $\lim_{n \to \infty} \left(1 - \frac{\log n}{n}\right) = 1$ . (It is true, I'm just not proving it here.)

I completely understand understand why you say that

Because it is completely false. Look at this.

**buckeye1973** · December 5th 2012, 04:28 PM

Originally Posted by Plato

I completely understand understand why you say that

Because it is completely false. Look at this.

No, no, no; you are misreading. I'm saying this. (Just the part inside the exponent.)

So, with that, is my proof okay?

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