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I'm trying to prove whether or not the following series converges:

$$\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}}{n+3}-\frac{\sqrt{n}}{n+4}\right).$$

I'm having some difficulty. I've been trying to compare it with something, but I'm having trouble making progress.

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$$\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}}{n+3}-\frac{\sqrt{n}}{n+4}\right)= \sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n-1}}{n+3}+\frac{\sqrt{n-1}}{n+3}-\frac{\sqrt{n}}{n+4}\right)= \sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n-1}}{n+3}\right)+ \sum_{n=1}^{\infty} \left( \frac{\sqrt{n-1}}{n+3}-\frac{\sqrt{n}}{n+4} \right)= \sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n-1}}{n+3}\right)= \sum_{n=1}^{\infty} \left(\frac{2}{(n+3)(\sqrt{n+1}+\sqrt{n-1})}\right) < \infty $$

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    $\begingroup$ Nice use of telescoping. $\endgroup$ – AndrewG Jan 6 '13 at 0:46
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Let $A=\dfrac{\sqrt{n+1}}{n+3}$ and $B=\dfrac{\sqrt{n}}{n+4}$. Multiply the $n$-th term by $\dfrac{A+B}{A+B}$. This is the usual "rationalizing the numerator" trick.

After some manipulation, we arrive at $$\frac{3n^2+15n+16}{(n+4)^2(n+3)\sqrt{n+1}+(n+3)^2(n+4)\sqrt{n}}.$$ The denominator is greater than $2n^{7/2}$, and after a while the numerator is less than $4n^2$.

So if $n$ is large enough, the $n$-th term is $\lt \dfrac{2}{n^{3/2}}$.

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$\frac{\sqrt{n+1}}{n+3}-\frac{\sqrt{n}}{n+4} ={1 \over \sqrt n}{\sqrt{1+1/n} \over 1+3/n} -{1 \over \sqrt n}{1 \over 1+4/n} ={1 \over \sqrt n}{(1+1/(2n)+O(1/n^2))(1-3/n+O(1/n^2)} -{1 \over \sqrt n}(1-4/n+O(1/n^2)) ={1 \over \sqrt n}(1-5/(2n)+O(1/n^2)) -{1 \over \sqrt n}(1-4/n+O(1/n^2)) ={1 \over \sqrt n}(-5/(2n)+4/n+O(1/n^2)) ={1 \over \sqrt n}(3/(2n)+O(1/n^2)) ={1 \over n^{3/2}}(3/2+O(1/n)) $.

Since each term is $O(1/n^{3/2})$, the sum converges.

The important thing is that the two terms in the summation are each ${1 \over \sqrt n}(1+O(1/n))$, so the ${1 \over \sqrt n}$ terms cancel, leaving only terms which are $O(1/n^{3/2})$, and the sum of these converges. As far as I am concerned, this would be a valid proof of convergence.

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  • $\begingroup$ Indeed it would, it is the right way to approach things. Surprised at the downvote, hadn't head of "big O" allergies. $\endgroup$ – André Nicolas Jan 6 '13 at 1:54
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$$\frac{\sqrt{n+1}}{n+3}-\frac{\sqrt n}{n+4}=\frac{(n+4)\sqrt{n+1}-(n+3)\sqrt n}{(n+3)(n+4)}$$ $$=\frac{(n+3)(\sqrt{n+1}-\sqrt n)}{(n+3)(n+4)}+\frac{\sqrt{n+1}}{(n+3)(n+4)}$$

$$=\frac{n+1-n}{(n+4)(\sqrt{n+1}+\sqrt n)}+\frac{\sqrt{n+1}}{(n+3)(n+4)}$$

Let $T_n=\frac 1{(n+4)(\sqrt{n+1}+\sqrt n)}$ and $S_n=\frac{\sqrt{n+1}}{(n+3)(n+4)}$

$\frac 1{(n+4)(\sqrt{n+1}+\sqrt n)}<\frac1{n(2\sqrt n)}=\frac1{2n^{\frac32}}$ or $\lim_{n\to \infty}\frac{\frac 1{(n+4)(\sqrt{n+1}+\sqrt n)}}{\frac1{n\sqrt n}}=2$

Using Comparison Test, with p-series (here $p=\frac32$), we can conclude $\sum_{n\to\infty}T_n $ is convergent.

Similarly, $\sum_{n\to\infty}S_n $ is convergent.

Applying $4$th property of this, $\sum_{n\to\infty}(S_n+T_n)$ is convergent.

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