2
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$$b_n=\frac{n^n}{(n+1)(n+2)\dots(n+n)}.$$ Now, there is this theorem for sequences that if $\lim_{n\to ∞}⁡ a_{n+1} /a_n =l$, $|l|<1$ then $\lim_{n\to ∞}⁡ a_n=0$. so, $\lim_{n\to ∞}⁡ b_{n+1} /b_n =e/4$ which is less than $1$, so $\lim_{n\to ∞}⁡ b_n$ should be equal to zero. But if I calculate the limit of $b_n$ as, $b_n=(n\cdot n\cdot n\cdots n)/((n+1)(n+2)...(n+n))$ I get $\lim_{n\to ∞}⁡ b_n=1/2$.

Something is definitely going wrong. Can someone point out my mistake,Please.

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    Your first way of calculation is right, and the second is wrong. Without knowing the details, I can't tell more. – 2017-01-14
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    We have to find the limit of the given sequence.What other details are required? – 2017-01-14
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    How exactly have you get to the $1/2$? I don't see a way to get there (and it is incorrect), so it would help if you described the $\lim_{n\to \infty}⁡ b_n=1/2$ part... – 2017-01-14
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    I mean, the details on how you arrived at $1\over2$. Without knowing those, I can't tell what's wrong, but there is definitely something wrong with that result. – 2017-01-14
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    Well, i wrote it as 1/((1+1/n)(1+2/n)..(2)) which i wrote to be equal to 1/2. I understand i am wrong here but. – 2017-01-14

2 Answers 2

1

I assume your argument looks like this:

$$b_n=\frac{n}{n+1} \frac{n}{n+2} ... \frac{n}{n+n}$$

Taking the limit as $n\to\infty$, each factor goes to $1$, eg. $\frac n {n+1}\to 1$, except the last one, since $\frac n {n+n}=\frac 1 2$. Thus the limit is $\frac 1 2$.

Well, what about the second-to-last factor? $\frac{n}{n+n-1}\to\frac 1 2$ as well. So is the limit $\frac 1 4$?

The basic problem is that the rule

$$\lim_{n\to\infty} a_n b_n = (\lim_{n\to\infty}a_n)(\lim_{n\to\infty}b_n)$$

Only applies when the number of factors doesn't depend on $n$.

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    ok. so how should i proceed? – 2017-01-14
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    @Paul Looking at the ratio $\frac {b_{n+1}} {b_n}$, as you did, is what I would have done too. And it worked, didn't it? – 2017-01-14
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    yeah,well. okay,then. thanks – 2017-01-14
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    @Jack M also third to last term also goes limiting to $\frac{1}{2}$ ? – 2017-01-14
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    @BAYMAX Yes. But the problem with the reasoning is that there basically is no "third to last term", and no "last term", either. When you say "the last term", you're talking about a different term for every $n$, because it's at first the first term, then the second, then the third... – 2017-01-14
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    @Jack M I agree but then how can we say what the limit is ? as you did limit was $\frac{1}{4}$ by assuming the last and second to last term .. but if we consider third to last term the limit will be $\frac{1}{8}$ so as we keep moving backwards from end we wll be getting more no. of $\frac{1}{2}$ , so can we conclude from there that limit is 0 ? – 2017-01-15
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    @BAYMAX No. There's just no way of justifying that rigorously. – 2017-01-15
1

One approach:

$$b_n=\frac{n^n}{(n+1)(n+2)\dots(n+n)}=\prod_{i=1}^n\frac{n}{n+i}$$

$$\implies \log b_n = n \cdot\left[ \frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)\right] \text{, where $f(x)=-\log(1+x)$}$$

So, the bracketed expression ought to tend to $\int_0^1f(x)\,dx=1-2\log2<0$

As such, $\log b_n = (1-2\log2)(n+o(1))\implies b_n=\left(\frac{e}{4}\right)^{n+o(1)}$