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EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post.

I tried expanding out the equation from the question and got $${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}$$

I then tried taking the ln of the equation which works out to $$\ln(1 - \frac{1}{2}) + \ln(1 - \frac{1}{4}) + ...$$

Here is my question. I used the rules of ln functions and took $ln(\frac{1}{1/2}) + ln(\frac{1}{\frac{1}{4}})$ + ...

Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then $$ln(L) = 0, where L = limit$$ $$L = e^0$$ $$L = 1$$

However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function $y = ln(x)$ to $y = x - 1$, as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve $y = x - 1$.

He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient.

David G. Stork
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statsguy21
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    How did you got to $\ln(\frac{1}{1/2}) + \ln(\frac{1}{\frac{1}{4}})$, this is wrong, the rule says that $\ln(a)-\ln(b)=\ln(a/b)$ not that $\ln(a-b)=\ln(a/b)$ – ℋolo Sep 06 '18 at 05:18
  • I see...so my concept was totally wrong in the first place. Should I edit it or should I just keep it up? – statsguy21 Sep 06 '18 at 05:23
  • Please look for similar questions before posting your new question. https://math.stackexchange.com/questions/2886753/limit-of-the-product-prod-k-1n-left-frac2k-12k-right, https://math.stackexchange.com/questions/1586773/evaluate-prod-frac-2k-1-2k, https://math.stackexchange.com/questions/93001/closed-form-expression-for-the-product-prod-limits-k-1n-left1-frac1, https://math.stackexchange.com/questions/2402508/convergence-point-of-frac-prod-k-1n-2k-1-prod-k-1n-2k?noredirect=1&lq=1 – JavaMan Sep 06 '18 at 05:26
  • @statsguy keep it up, or if edit just explain you made a mistake. It is better to show effort in the post(something that you clearly did) – ℋolo Sep 06 '18 at 05:28
  • @JavaMan I don't think this is a duplicate, this is question about intuition behind the linear approximation, and not the product itself(Or so I think) – ℋolo Sep 06 '18 at 05:28
  • @JavaMan yes I'm sure there are much better ways to find the answer, but I'm looking for more of the intuition behind this approach of using the tangent to find the sum to infinity of a series. – statsguy21 Sep 06 '18 at 05:30
  • @Holo I'll just keep it up then, if I edit it the whole post not just that part but all the other calculations won't make sense anymore – statsguy21 Sep 06 '18 at 05:32
  • If your real question does not involve the question you put in the title, then change your title. Put some effort into your posts, and you'll get quality responses back. – JavaMan Sep 06 '18 at 05:32
  • There is nothing wrong with the title, I was trying to find the limit of the sequence. But I have edited the post as well to reflect that I was wrong with the ln rules. – statsguy21 Sep 06 '18 at 05:35

3 Answers3

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A completely different approach is to write $${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}=\frac {(2n)!}{(2^nn!)^2}$$ because you can divide a $2$ out of each term on the bottom and get $n!$ and you can multiply top and bottom by the bottom. Now feed it to Stirling $$a_n\approx\frac{(2n)^{2n}e^{2n}}{e^{2n}2^{2n}n^{2n}\sqrt{\pi n}}=\frac 1{\sqrt{\pi n}}\to 0$$

Ross Millikan
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  • nice! Why can't we use the justification that each ratio individually is less than 1 and since there are infinite terms, the product should tend to zero? (sorry, am from a non-mathematical background) – dineshdileep Sep 06 '18 at 05:40
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    @dineshdileep That's not true in general. Example $\prod _2^\infty (1-2/(n(n+1))) = 1/3$. – xbh Sep 06 '18 at 05:45
  • @dineshdileep No, you can't say that. A well-known counterexample is $\lim_{n \to \infty}(1-\frac{1}{n})^n = \frac{1}{e}$ – ab123 Sep 06 '18 at 05:46
  • Powerful as always. – Mikasa Sep 06 '18 at 05:53
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As you say, $$\ln a_n=\sum_{k=1}^n\ln\left(1-\frac1{2k}\right).$$ But as $x\to0$, $$\ln(1-x)=-x+O(x^2).$$ Therefore $$\ln a_n=-\sum_{k=1}^n\frac1{2k}+O\left(\sum_{k=1}^n\frac1{k^2}\right).$$ As the series $\sum_1^\infty1/k$ diverges and $\sum_1^\infty1/k^2$ converges, then $\ln a_n\to-\infty$, and so $a_n\to0$.

Angina Seng
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1

We can prove that $${a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}<\frac{1}{\sqrt{2n+1}}.$$ Let $b_n=\frac{2}{3} .\frac{4}{5} . \frac{6}{7} ...\frac{2n}{2n+1}$, then it is easy to see that $a_n<b_n$. So $$a_n^2<a_nb_n=\frac{1}{2n+1}.$$

You limit is $$\lim_{n\to \infty}\frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}=0.$$

Riemann
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