1
$\begingroup$

I have an objective type question:-
If $$a_n=\left(1-\frac{1}{\sqrt{2}}\right)\ldots\left(1-\frac{1}{\sqrt{n+1}}\right)$$ then $\lim_{n\to\infty}a_n=?$:-
A)$0$
B)limit does not exist
C)$\frac{1}{\sqrt\pi}$
D)1
My approach :-As the product contains $1/\sqrt2$ so the overall product would be irrational,and as product is converging also so by option elimination answer would be $$C) \frac{1}{\sqrt \pi}$$,as it is the only irrational number in options.
Now i want to ask is that is my solution right? If it is right then what is the proper method to actually solve the question? but if it is wrong then what is the right solution?
Suddenly a doubt is also arising that whether 0 is an irrational number?

$\endgroup$

marked as duplicate by quid Feb 25 at 14:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Is the product of $1/\sqrt{2}$ with itself still irrational ? $\endgroup$ – mwoua Apr 12 '16 at 10:33
  • $\begingroup$ It is rational,That means i was wrong ......ok so put the right solution please in the answers. $\endgroup$ – Mayank Deora Apr 12 '16 at 10:39
  • $\begingroup$ Note that a limit of irrational numbers can br rational, consider $\lim_{n \to \infty} \frac {1}{n \pi} $. To find the limit, look at all square terms. $\endgroup$ – Wouter Rienks Apr 12 '16 at 10:43
2
$\begingroup$

A simple estimate yields $$0 < a_n = \left( 1- \frac{1}{\sqrt{2}} \right) \cdots \left( 1- \frac{1}{\sqrt{n+1}} \right) < \left( 1- \frac{1}{\sqrt{n+1}} \right)^n = b_n.$$ The series$$\sum \limits_{n=1}^\infty b_n$$ converges by the comparison test. Thus $b_n \to 0$ and $a_n \to 0$.

$\endgroup$
  • $\begingroup$ Sorry for all the edits. But this should work. $\endgroup$ – Matias Heikkilä Apr 12 '16 at 10:45
  • $\begingroup$ You could use the theorem about the convergence of bounded monotone sequences directly... The series thing is a bit of an overkill, but the point stands. $\endgroup$ – Matias Heikkilä Apr 12 '16 at 10:49
  • $\begingroup$ I liked your first two lines & so i used it myself by applying a formula of $1^\infty$ giving $e^{-\infty}=0$ and since $a_n$ lies between 0 & $b_n$ it would also converge to 0, is my approach correct? $\endgroup$ – Mayank Deora Apr 12 '16 at 11:05
1
$\begingroup$

$\displaystyle \ln a_n = \sum_{i=2}^{n+1} \ln(1-\frac{1}{\sqrt{i}})$ and $\displaystyle \ln(1-\frac{1}{\sqrt{i}})\sim -\frac{1}{\sqrt{i}}$

Thefore $\sum_{i=2}^{n+1} \ln(1-\frac{1}{\sqrt{i}})$ diverges to $-\infty$

Hence $\ln a_n\to -\infty$

Hence $a_n\to 0$

$\endgroup$
0
$\begingroup$

\begin{align}\lim_{n\to\infty}a_n&=\left(1-\frac1{\sqrt2}\right)\ldots\left(1-\frac1{\sqrt{n+1}}\right)\le\lim_{n\to\infty}\left(1-\frac1{\sqrt{n+1}}\right)^{n+1}\!\!\Big/\left(1-\frac1{\sqrt{n+1}}\right)\\&=\lim_{x\to\infty}\left(1-\frac1{x}\right)^{x^2}\!\!\Big/\ 1=\lim_{x\to\infty}\left(\left(1-\frac1{x}\right)^x\right)^x=\left(\frac1e\right)^{\infty}=0\end{align}

$\endgroup$
0
$\begingroup$

The real issue here is that the natural logarithm of $(1-\frac 1{\sqrt n})$ is roughly $-\frac 1{\sqrt n}$ (in the sense that the ratio goes to $1$). So the sum of the logarithms of these terms is roughly the partial sum of $-\sum \frac 1{\sqrt n}$. Since the latter diverges, the logarithms go to $-\infty$, so the limit of $a_n$ is $0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.