Find the sum of the limit $$\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{2^k}\sum_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}\right)$$

My try: since $$\sum_{i=0}^{2^{k-1}-1}\ln{\left(\dfrac{2^k+2+2i}{2^k+1+2i}\right)}=\sum_{i=0}^{2^{k-1}-1}\left(\ln{(2^k+2+2i)}-\ln{(2^k+1+2i)}\right)$$

My friend tells me this sum has an analytical solution. But I can't find it. Thank you.


Note that $$ \begin{align} &\prod_{i=0}^{2^{k-1}-1}\frac{2^k+2+2i}{2^k+1+2i}\\ &=\frac{2^k+2}{2^k+1}\frac{2^k+4}{2^k+3}\frac{2^k+6}{2^k+5}\cdots\frac{2^k+2^k}{2^k+2^k-1}\\ &=\frac{2^k+2}{2^k+1}\color{#A0A0A0}{\frac{2^k+2}{2^k+2}}\frac{2^k+4}{2^k+3}\color{#A0A0A0}{\frac{2^k+4}{2^k+4}}\frac{2^k+6}{2^k+5}\color{#A0A0A0}{\frac{2^k+6}{2^k+6}}\cdots\frac{2^k+2^k}{2^k+2^k-1}\color{#A0A0A0}{\frac{2^k+2^k}{2^k+2^k}}\\ &=\frac{\left(2^{2^{k-1}}2^k!/2^{k-1}!\right)^2}{2^{k+1}!/2^k!}\\ &=\left(\frac1{4^{2^{k-1}}}\left.\binom{2^k}{2^{k-1}}\right)\middle/\left(\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\right)\right.\tag{1} \end{align} $$ Furthermore, it is easy to show by induction that $$ \begin{align} \prod_{k=1}^n\left(\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\right)^{1/2^{k+1}} &=\frac1{2^n}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\tag{2} \end{align} $$ Therefore, let $$ f(k)=\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\tag{3} $$ Then $$ \begin{align} &\sum_{k=1}^n\left(\frac1{2^k}\sum_{i=0}^{2^{k-1}-1}\log\left(\frac{2^k+2+2i}{2^k+1+2i}\right)\right)\\ &=\sum_{k=1}^n\frac1{2^k}\Big(\log(f(k-1))-\log(f(k))\Big)\\ &=\sum_{k=0}^{n-1}\frac1{2^{k+1}}\log(f(k))-\sum_{k=1}^n\frac1{2^k}\log(f(k))\\ &=\frac12\log(f(0))-\frac1{2^n}\log(f(n))-\sum_{k=1}^{n-1}\frac1{2^{k+1}}\log(f(k))\\ &=-\frac12\log(2)-\frac1{2^n}\log(f(n))-\log\left(\frac1{2^n}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\right)\tag{4} \end{align} $$ Since Stirling says $$ \lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e\tag{5} $$ we get $$ \begin{align} &\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{2^k}\sum_{i=0}^{2^{k-1}-1}\log\left(\frac{2^k+2+2i}{2^k+1+2i}\right)\right)\\ &=-\frac12\log(2)-\lim_{n\to\infty}\log\left(\frac2{2^{n+1}}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\right)\\ &=-\frac12\log(2)-\log\left(\frac{\sqrt2}{e}\right)\\[6pt] &=1-\log(2)\tag{6} \end{align} $$

  • $\begingroup$ Great answer (+1) $\endgroup$ – user 1591719 Dec 31 '13 at 21:01
  • $\begingroup$ It's nice too!Thank you $\endgroup$ – math110 Jan 1 '14 at 4:06



and $\ln(1+u)=u-\tfrac12 u^2+\tfrac13u^3\mp...$

Probably more tricky than that. Write down the sums up to $n=4$ in your transformation of the logarithms and consider that for example

$\ln(14)=\ln(2)+\ln(7)$ and $\ln(28)=2\ln(2)+\ln(7)$.

I'd expect that the logarithms of the odd numbers cancel in the long run, however I do not see what the buildup of the $\ln(2)$ terms leads to.

Expanded: Define $$A_k=\sum_{i=0}^{2^{k-1}-1}\ln(2^k+2i+2)$$ for the sum of logs of even numbers between $2^k+1$ and $2^{k+1}$ and $$B_k=\sum_{i=0}^{2^{k-1}-1}\ln(2^k+2i+1)$$ for the sum of logs of odd numbers in the same segment. Then the task is to determine the limit for

$$\sum_{k=1}^n \frac{A_k-B_k}{2^k}$$

But we also have the recursion that the even numbers can be divided by $2$ and thus mapped to the previous interval, $$A_{k+1}=2^k\ln(2)+A_k+B_k,$$ so the big sum reduces to

$$\sum_{k=1}^n \frac{2^k\ln2+2A_k-A_{k+1}}{2^k}=n\ln(2)+A_1-\frac{A_{n+1}}{2^n}$$

and for the last term we can finally reach back to the Riemann sum techniques.

I may have overlooked something, ... according to my quick'n'dirty calculation the result could be just simply $$1-\ln(2).$$

  • $\begingroup$ First,Thank you,and then?I don't think this is follow not work $\endgroup$ – math110 Dec 31 '13 at 8:16
  • $\begingroup$ Yes, the terms for small k would render this useless. It will probably lead to the consideration of geometric series for the powers of $\frac12$. $\endgroup$ – Lutz Lehmann Dec 31 '13 at 8:22

studying the terms of the series more carefully (after my initial crass misreading!) it occurred to me that we are dealing here with an assertion about geometric means.

let $$s_n=\sum_{k=1}^{n}\frac{\sum\limits_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}}{2^k} $$ and define $g_n$ to be the geometric mean of the $2^n$ consecutive integers $2^n+1, 2^n+2,\dots,2^{n+1}$. then we have (something like): $$ e^{s_n} = \frac{2^n}{g_n}$$

to find a more precise value we may write: $$ g_n^{2^n}= \frac{2^{n+1}!}{2^n!} $$ to which we may apply Stirling's approximation for the gamma function. I am happy to leave the more intricate details of the calculation to others less prone to silly mistakes. however a quick back-of-the envelope calculation suggests that: $$\lim_{n\to\infty} s_n = \frac{e}4 $$ note to self. 1)forgot to take logs at the end (pretend I ran out of envelope) 2) forgot 1st numerator doesn't cancel!

  • 1
    $\begingroup$ I get $g_0=2$ and thus $e^{s_n}=2^n\frac{g_0}{g_n}=\frac{2^{n+1}}{g_n}$, which would lead to the limit $\frac e2$ for this series of exponentials and $\lim s_n=1-\ln(2)$. Happy new year. $\endgroup$ – Lutz Lehmann Jan 1 '14 at 8:28
  • $\begingroup$ But of the three parametrizations presented, I like your use of the geometric mean the best. It allows for a short and lucid presentation of the solution. $\endgroup$ – Lutz Lehmann Jan 1 '14 at 9:15
  • $\begingroup$ @Lutzl your answer was correct(+1). I forgot the 1st numerator. a typical glitch. from an earlier comment you made I think you had seen the cancellation from playing with the formula. even after I saw the GM connection I struggled for a while before realizing the "squared numerator" is compensated for by the extra power of two in the root taken. wish you the best in 2014 ;-) $\endgroup$ – David Holden Jan 1 '14 at 9:23

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