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Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.

My attempt is

$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish to find $V_n $ which is less than (or greater than $U_n $) such that its convergence/divergence is known so that I can find the behaviour of $U_n $

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2 Answers 2

up vote 3 down vote accepted

Here, $\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq \frac{1}{n^2}$. For n=1, it is true, and then the denominator of given series increases more rapidly than numerator and this rate is greater than the rate in $1/n^2$, so the given series is bounded by $1/n^2$ and hence $\sum_{1}^{\infty}\frac{\sqrt{n+1}-1}{(n+2)^3-1} \leq (\sum_{1}^{\infty}\frac{1}{n^2}=\pi^2/6)$. Thus the series converges.

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The normal way of finding a larger series is decrease the denominator and/or increase the numerator. Here, we have found larger series by inspection. Can it be done by normal way (decreasing denominator etc.)? –  rajendra bakre Jul 3 '12 at 16:55
    
Decrease denominator from $(n+2)^3-1$ to $n^3$ and increase numerator from $\sqrt{n+1}-1$ to $n$ and then you are done. –  Aang Jul 3 '12 at 17:15

Hint: $\sqrt{n+1}-1\sim\sqrt n$ and $(n+2)^3-1\sim n^3$ (when $n\to +\infty$). Since the series have non-negative terms, we just have to deal with the convergence of $\sum\limits_{n\geq 1}\frac{\sqrt n}{n^3}$.

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