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Lets formulate the following series of fractions a_1/b_1 ,a_2=a_1+1/ b_2=a_1+b_1+2, a_3=a_2+1/b_3=a_2+b_2+2 and so on.If a_1=2 and b_1=4 we have the following series of fractions.2/4+3/8+4/13+5/19+6/26+7/34+..... How do we prove the infinite sum of these fractions diverges?

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You're notation is confusing. Are you defining $$a_n = a_{n-1} + 1 \qquad b_n = a_{n-1} + b_{n-1} + 2 $$and considering the fractions $a_n / b_n$? – Hurkyl Dec 15 '12 at 2:58
up vote 2 down vote accepted

The $n^{th}$ term in the series is given by $x_n = \dfrac{n+1}{(n+1)(n+4)/2-1} = \dfrac{2(n+1)}{n^2 + 5n + 2}$. Note that $$\dfrac{(n+1)}{n^2 + 5n + 2} \geq \dfrac12 \dfrac1{n+1}$$ Hence, $x_n \geq \dfrac1{n+1}$. Hence, we get that $$\sum_{n=1}^{\infty} x_n \geq \sum_{n=1}^{\infty} \dfrac1{n+1}$$ Hence, $\displaystyle \sum_{n=1}^{\infty} x_n$ diverges.

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Assuming you mean the recurrence in my comment, we can solve the recursion exactly using standard techniques for solving difference equations, or writing out the first few terms and seeing a pattern.

A shorter argument would observe that $a_n$ grows linearly in $n$. $b_n$ grows like the summation of something that grows linearly, and so $b_n$ should grow quadratically.

Therefore, $a_n / b_n$ shrinks like $1/n$, and so the sum should diverge.

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