# Limiting value of a sequence when n tends to infinity [duplicate]

Q) Let, $$a_{n} \;=\; \left ( 1-\frac{1}{\sqrt{2}} \right ) ... \left ( 1- \frac{1}{\sqrt{n+1}} \right )$$ , $$n \geq 1$$. Then $$\lim_{n\rightarrow \infty } a_{n}$$

(A) equals $$1$$

(B) does not exist

(C) equals $$\frac{1}{\sqrt{\pi }}$$

(D) equals $$0$$

My Approach :- I am not getting a particular direction or any procedure to simplify $$a_{n}$$ and find its value when n tends to infinity.
So, I tried like this simple way to substitute values and trying to find the limiting value :-
$$\left ( 1-\frac{1}{\sqrt{1+1}} \right ) * \left ( 1-\frac{1}{\sqrt{2+1}} \right )*\left ( 1-\frac{1}{\sqrt{3+1}} \right )*\left ( 1-\frac{1}{\sqrt{4+1}} \right )*\left ( 1-\frac{1}{\sqrt{5+1}} \right )*\left ( 1-\frac{1}{\sqrt{6+1}} \right )*\left ( 1-\frac{1}{\sqrt{7+1}} \right )*\left ( 1-\frac{1}{\sqrt{8+1}} \right )*.........*\left ( 1-\frac{1}{\sqrt{n+1}} \right )$$

=$$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$$ =0.009*...

So, here value is tending to zero. I think option $$(D)$$ is correct.
I have tried like this
$$\left ( \frac{\sqrt{2}-1}{\sqrt{2}} \right )*\left ( \frac{\sqrt{3}-1}{\sqrt{3}} \right )*\left ( \frac{\sqrt{4}-1}{\sqrt{4}} \right )*.......\left ( \frac{\sqrt{(n+1)}-1}{\sqrt{n+1}} \right )$$
= $$\left ( \frac{(\sqrt{2}-1)*(\sqrt{3}-1)*(\sqrt{4}-1)*.......*(\sqrt{n+1}-1)}{{\sqrt{(n+1)!}}} \right )$$
Now, again I stuck how to simplify further and find the value for which $$a_{n}$$ converges when $$n$$ tends to infinity . Please help if there is any procedure to solve this question.

## marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, PedroFeb 24 at 11:56

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.

• Using the inequality $1+x \leq e^x$ which is true for all $x \in \mathbb{R}$, we have $$a_n \leq \exp\left\{ -\sum_{k=2}^{n} \frac{1}{\sqrt{k}} \right\} \leq C \exp\left\{ -\int_{0}^{n} \frac{\mathrm{d}x}{\sqrt{x}} \right\} = C e^{-2\sqrt{n}}$$ for some constant $C > 0$. From this, we can conclude $a_n \to 0$. A more natural approach is to take log and utilize Taylor approximation, which in turn yields $a_n = e^{-2\sqrt{n} + \mathcal{O}(1)}$ as $n\to\infty$. – Sangchul Lee Feb 24 at 9:12
• What would you do if it were $$\left(1-\frac12\right)\left(1-\frac13\right)\cdots\left(1-\frac1{n+1}\right)?$$ – Lord Shark the Unknown Feb 24 at 9:13
• – Sil Feb 24 at 9:32

## 3 Answers

The hint: $$0

• Michael.In 2 lines! :) – Peter Szilas Feb 24 at 9:28

As we multiply positive factors below $$1$$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $$a_1<1$$, we also rule out A.

If we multiply $$a_n$$ by $$b_n:=\left(1+\frac1{\sqrt 2}\right)\cdots \left(1+\frac1{\sqrt {n+1}}\right)$$, note that the product of corresponding factors is $$\left(1-\frac1{\sqrt {k}}\right)\left(1+\frac1{\sqrt {k}}\right)=1-\frac1k<1$$, hence $$a_nb_n<1$$. On the other hand, by expanding the product and dropping lots of positive terms $$b_n\ge 1+\frac1{\sqrt 2}+\frac1{\sqrt 3}+\ldots+\frac1{\sqrt {n+1}}$$ so that $$b_n$$ gets arbitrarily large. We conclude that $$a_n$$ gets arbitrarily small positive. In other words $$a_n\to 0$$.

• Hagen.Very nice. – Peter Szilas Feb 24 at 9:27

Here is a simple approach to rule out the wrong options and therefore find the correct one. I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.

The numbers $$1-1/\sqrt{k}$$ are all in $$(0,1)$$, so your sequence is positive and strictly decreasing. Therefore it has a limit. You have already computed enough terms to rule out the limits $$1$$ and $$1/\sqrt{\pi}$$. Remember that it's decreasing. The only remaining option is that the limit is zero.