1

Let $$ a_n = \left( 1 - \frac{1}{\sqrt 2} \right) \dotsm \left( 1 - \frac{1}{\sqrt {n+1}} \right), \quad n\geq 1. $$ Then $\lim_{n \to \infty} a_n$

  1. is $1$,
  2. is $0$,
  3. does not exist,
  4. is $\frac{1}{\sqrt\pi}$.

I tried this using the sandwich theorem as $$ 0 \leq \sum_{r=1}^{n} \frac{-1}{\sqrt r+1} \leq \frac{-n}{\sqrt {n+2}} $$ which goes to zero. I am not sure though.

Jendrik Stelzner
  • 16,401
Sophie Clad
  • 2,291
  • In general, what is $a_n$? It is not clear from $\left(1-\frac1{\sqrt2}\right)\dots\left(1-\frac1{n+1} \vphantom{\frac1{\sqrt2}}\right)$ – robjohn Jan 01 '16 at 08:34
  • a $\sqrt{..}$ is missing? – Dr. Sonnhard Graubner Jan 01 '16 at 08:36
  • @Dr.SonnhardGraubner: you mean $\left(1-\frac1{\sqrt2}\right)\dots\left(1-\frac1{\sqrt{n+1}}\right)$? – robjohn Jan 01 '16 at 08:37
  • @robjohn yeah edited it – Sophie Clad Jan 01 '16 at 08:42
  • $\frac n{n+2}\to1$, so the sandwich theorem doesn't apply – robjohn Jan 01 '16 at 08:47
  • @robjohn edited it – Sophie Clad Jan 01 '16 at 08:49
  • @SophieClad: It seems you are trying to use the theorem that for $0\le a_n\lt1$ and $\sum\limits_{n=1}^\infty a_n=\infty$ then $\prod\limits_{n=1}^\infty\left(1-a_n\right)=0$. Then $\sum\limits_{k=1}^n\frac1{\sqrt{k+1}}\ge\frac{n}{\sqrt{n+1}}$ and since $\frac{n}{\sqrt{n+1}}\to \infty$, you get $\prod\limits_{n=1}^\infty\left(1-a_n\right)=0$. If this is what you are trying to show, then you are right. (+1) – robjohn Jan 01 '16 at 08:59

2 Answers2

5

Using $$ a_n=\left(1-\frac1{\sqrt2}\right)\cdots\left(1-\frac1{\sqrt{n+1}}\right) $$ then, because $$ \left(1-\frac1{\sqrt{n+1}}\right)\le\left(1-\frac1{n+1}\right) $$ and $$ \begin{align} \left(1-\frac12\right)\cdots\left(1-\frac1{n+1}\right) &=\frac12\cdot\frac23\cdot\frac34\cdots\frac{n}{n+1}\\ &=\frac1{n+1} \end{align} $$ we have $$ a_n\le\frac1{n+1} $$

robjohn
  • 345,667
1

$(1-\dfrac{1}{\sqrt {n+1}})^n\geq a_n=(1-\dfrac{1}{\sqrt 2})...(1-\dfrac{1}{\sqrt {n+1}})\geq (1-\dfrac{1}{\sqrt 2})^n$

As $(1-\dfrac{1}{\sqrt 2})<1\implies (1-\dfrac{1}{\sqrt 2})^n\to 0$

Also $(1-\dfrac{1}{\sqrt {n+1}})^n\approx\dfrac{1}{e^{\sqrt n}}\to 0$

Hence $\lim a_n\to 0$

Learnmore
  • 31,062