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If $a_{1},a_{2},a_{3},.....,a_{n}$ are n terms of series such that $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}\;,n\geq 1, n\in \mathbb{N}$$

Then $\displaystyle n^4\lim_{n\rightarrow \infty}\prod^{n}_{r=1}a_{r} = $

$\bf{My\; Try::}$ Given $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}=2+\frac{4}{n}$$

We can Write it as $$\frac{n+1}{a_{n+1}}-\frac{n}{a_{n}} = \frac{1}{a_{n}}+2+\frac{4}{n}$$

Now Put $n=1,2,3,4,.......n$ and adding, We get

$$\frac{n+1}{a_{n+1}}-\frac{1}{a_{n}} = \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+......+\frac{1}{a_{n}}\right)+2n+4\left(\frac{1}{1}+\frac{1}{2}+.....+\frac{1}{n}\right)$$

Now How can I solve after that, Help me

Thanks

Community
  • 1
juantheron
  • 53,015

2 Answers2

2

HINT :

We can write $$\frac{n+1}{a_{n+1}}-\frac{n-1}{a_{n}} = \frac{2(n+2)}{n}$$ as $$b_{n+1}-b_n=n+2$$ where $$b_n=\frac{\binom n2}{a_n}$$

mathlove
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Define $$b_n := \frac{n(n-1)}{a_n}$$ Then your recurrence reduces to $$b_{n+1}-b_n=2n+4$$ So, $$b_n = b_1 + 2\sum_{i=1}^{n-1} i + 4\sum_{i=1}^{n-1} 1 = b_1 + n(n-1)+4(n-1)=b_1+n^2+3n-4$$ $$\prod_{i=1}^n a_i = \prod_{i=1}^n \frac{n(n-1)}{b_i} = \prod_{i=1}^n \frac{n^2-n}{b_1+n^2+3n-4}$$

Now from the value of $b_1=0$ for nonzero $a_1$ so I think there might be some telescoping here.

$$\prod_{i=1}^n \frac{n}{n+4} = \frac{1}{\color{red}{5}}\frac{2}{\color{red}{6}}\frac{3}{\color{red}{7}}\frac{4}{\color{red}{8}}\color{red}{\frac{\color{red}{5}}{9}\frac{{6}}{10}\dots\frac{n-4}{n}}\frac{\color{red}{n-3}}{n+1}\frac{\color{red}{n-2}}{n+2}\frac{\color{red}{n-1}}{n+3}\frac{\color{red}{n}}{n+4}=\frac{4!}{(n+1)(n+2)(n+3)(n+4)}$$

So the limit is (assuming you meant $n^4$ inside the limit) $$\lim_{n\to\infty} n^4 \frac{4!}{(n+1)(n+2)(n+3)(n+4)} = \fbox{24}$$

Aritra Das
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  • All we know about $a_1$ is that $a_1\not= 0$. Especially, we don't get that $a_1=1/5$, and so $a_1$ has to be included in the answer, I think. – mathlove May 13 '16 at 07:03
  • @mathlove Uhh, but using my (and your) definition of ${b_n}$, $b_1$ is zero for all $a_1$. Except of course $0$. – Aritra Das May 13 '16 at 07:05
  • @mathlove Now that I'm thinking about it, I seem to be trapped in some sort of loop of logic. – Aritra Das May 13 '16 at 07:10
  • $a_1$ is not determined by the original equation. Setting $n=1$ gives us nothing about $a_1$ other than $a_1\not=0$. – mathlove May 13 '16 at 07:12
  • @mathlove Yes, I realised that. The recurrence only states how the terms are related to each other. We need the starting point to uniquely determine the sequence and hence the limit. – Aritra Das May 13 '16 at 07:17