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In the sequence $\frac{1}{2}, \frac{5}{3}, \frac{11}{8}, \ldots$ the $(n+1)$ st term is the sum of the numerator and the denominator of the $n$ th term. The numerator of the $(n+1)$ st term is the sum of the denominators of the $(n+1)$ st term and the $n$ th term. Find the limit of this sequence.

In this problems isn't ever term after the first of form $a+\frac{b}{c}$ where $a,b,c\in\mathbb{N}$. So as the each term is greater than $1$ (other than the first term of course), this sequence should tend to infinity. Then how are we suppose to find its limit?

Bill Dubuque
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Abhinandan Saha
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    Include your attempts. Otherwise this question will get closed. – vitamin d May 03 '21 at 10:46
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    Is there a typo in the first line of your question? Should it read the denominator of the $(n+1)$-st term is the sum of the numerator and the denominator of the $n$-th term? – user2661923 May 03 '21 at 11:08
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    Hint: Forget the math for the moment, and attack the problem with your intuition. Suppose that it converges to a value $L$. That is, it is converging to something that looks like $\frac{Lk}{k}$. If $a_n = \frac{Lk}{k} = a_{n+1} = \frac{(L+2)k}{(L+1)k}$, can you solve for $L$? This doesn't complete the problem, there would still be a great deal to do, such as manually computing $a_1, a_2, \cdots, a_{10}$ and looking for a pattern. However, it is at least a beginning. – user2661923 May 03 '21 at 11:23
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    Have you tried calculating the first few terms? That is often a quick way to see what might be happening: in this case you will see that the fractions converge quickly to something that looks very like $\sqrt 2$. – WA Don May 03 '21 at 11:48
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    Just to be clear, the problem is not limited to finding the limit, but also includes proving that the limit exists. That is, you have to prove that the sequence is convergent. Note that the algebraic manipulation re my previous comment is invalid if the sequence is divergent. This means that you can not use that type of analysis to prove that the limit converges. While you can use that type of analysis to guide your thinking, the formal proof of convergence must precede the algebraic manipulation. – user2661923 May 03 '21 at 13:59
  • The reasoning presented in this question and its answers can be directly applied here. As commenters have pointed out, the sequence converges to $\sqrt{2}$, and it is obvious that the result is independent of the initial term. Given the current EoQS policy and how long ago the question was posted, I doubt anyone will be answering. Shall we close as duplicate? – user3733558 Jun 20 '21 at 11:51
  • Does this answer your question? Approximation to $ \sqrt{2}$ – user3733558 Jun 20 '21 at 11:52

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