Infinite series converging to natural log

[zodiak770]

Prove that sum 1/2^(n+1)*n/(n+1), from n=0 to infinity, converges to 1 - log(2), where log stands for the natural logarithm.

I know that the Taylor series for log(x) about x=1 is sum (-1)^(n+1)*(x-1)^n/n, but I don't see how these two statements are consistent.

Thanks for any pointers!

 

[ystael]

The [tex]n(n + 1)[/tex] in the denominator is a clue that you should try expressing each term of the series as the difference of two fractions.

 

[zodiak770]

sorry, what n(n+1) in the denominator? Also, there was a typo in my original post which I have since corrected--the first sum had terms of the form 2^(n+1)*n/(n+1), and it should have been 1/2^(n+1)*n/(n+1).

 

[ystael]

Ah -- your expression is ambiguously parenthesized and so I read what I expected to see, not what you intended. Do you mean the sum is [tex]\sum_{n\geq 1} \frac1{2^{n+1}} \frac{n}{n + 1}} = \sum_{n\geq 1} \frac{n}{2^{n+1}(n + 1)}[/tex]? If so, to write it the way you did you should parenthesize: (1/2^(n+1))(n/(n + 1)).

 

[zodiak770]

yes--that's what I meant. Sorry for not making it more readable.

 

[ystael]

OK, here's a hint: you should, in fact, be thinking about the Taylor series which you wrote above, [tex]\log(1 + u) = \sum_{n\geq 1} \frac{(-1)^{n+1}u^n}{n}[/tex] -- but the correct value of [tex]u[/tex] is not [tex]1[/tex].

 

[zodiak770]

well, my first thought is express 1 - log(2) = log(e/2)=log( 1+(e-2)/2 ), ie, letting u be (e-2)/2, and then we end up with the = sum { (-1)^(n+1)*((e-2)/2)^n/n } from n=0 to infinity... is this the right track?

 

[ystael]

You have one good idea and one bad idea in there. The bad idea is the one that introduces something that isn't in the series you're trying to mimic; the good idea is the one that introduces something that is in it.

 

[inline]

yes--that's what I meant. Sorry for not making it more readable.

Divide the series into two fractions:
[tex] S = \sum_{n=0}^\infty \frac 1 {2^{n+1}} - \sum_{n=0}^\infty \frac 1 {2^{n+1}(n+1)} [/tex]
The first series is a geometrical progression, the second can be represented as [tex]\int_0^{1/2}\sum_{n=0}^\infty x^{n} dx[/tex].

 

[Dick]

Divide the series into two fractions:
[tex] S = \sum_{n=0}^\infty \frac 1 {2^{n+1}} - \sum_{n=0}^\infty \frac 1 {2^{n+1}(n+1)} [/tex]
The first series is a geometrical progression, the second can be represented as [tex]\int_0^{1/2}\sum_{n=0}^\infty x^{n} dx[/tex].

You might also note that the second sum is a sum you can get out of your expansion of log(x-1) for a particular value of x.

 

[zodiak770]

Brilliant--thank you guys so much. This problem had been gnawing at me for almost a week. In case any of you were wondering, this wasn't a homework problem--it was motivated by a google interview question. Check it out: http://mathoverflow.net/questions/17960/google-question-in-a-country-in-which-people-only-want-boys