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The general term $a_n$ for a sequence is given below. Test if the sequence converges and find the limit if it exists.

$$a_n = \sqrt{n^2 + 5n} - \sqrt{n^2 + 2n}$$

Now, I know from plotting this sequence using software that it does indeed converge to $1.5$.

enter image description here

My problem is that I'm having trouble showing it analytically. The root test for sequence convergence doesn't lead anywhere, so the most likely successful approach appears to be the ratio test, where to start I am required to evaluate $$ L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|, $$ but this turns very messy very quickly: $$ L = \lim_{n \to \infty} \left|\frac{ \sqrt{(n+1)^2 + 5(n+1)} - \sqrt{(n+1)^2 + 2(n+1)}}{\sqrt{n^2 + 5n} - \sqrt{n^2 + 2n}}\right|, $$ $$ L = \lim_{n \to \infty} \left|\frac{ \sqrt{(n+1)^2 + 5(n+1)} - \sqrt{(n+1)^2 + 2(n+1)}}{\sqrt{n^2 + 5n} - \sqrt{n^2 + 2n}} \frac{\sqrt{n^2 + 5n} + \sqrt{n^2 + 2n}}{\sqrt{n^2 + 5n} + \sqrt{n^2 + 2n}}\right|. $$ etc.

I've made various attempts from this point onwards to get to a point where I could evaluate the limit, but it doesn't lead to me getting a value for $L$ (and then confirming that the sequence does indeed converge to $1.5$). Am I missing something obvious here?

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  • $\begingroup$ Excuse me, but surd is it a software? $\endgroup$ – Sebastiano Feb 10 '20 at 23:24
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    $\begingroup$ surd means $\sqrt{}$ $\endgroup$ – J. W. Tanner Feb 10 '20 at 23:36
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    $\begingroup$ @J.W.Tanner Thank you very much and excuse me for my bad english. $\endgroup$ – Sebastiano Feb 10 '20 at 23:40
  • $\begingroup$ The root and ratio tests may tell you if a sequence or series converges or diverges. But they cannot tell you what the series converges to. (Although a convergent series would mean that the sequence converges to 0.) $\endgroup$ – Teepeemm Feb 11 '20 at 18:15
  • $\begingroup$ For large $n$ this is not going to be far away from $\sqrt{n^2 + 5n +\frac{25}{4} } - \sqrt{n^2 + 2n+1}$ $\endgroup$ – Henry Feb 11 '20 at 18:16
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Multiply numerator and denominator by $\sqrt {n^{2}+5n}+\sqrt {n^{2}+2n}$ to see that $a_n=\frac {3n} {\sqrt {n^{2}+5n}+\sqrt {n^{2}+2n}}$. Divide all the terms by $n$ to see that the limit is $\frac 3 2$.

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  • $\begingroup$ Perfect, thank you very much. In fact, I had failed to recognise that $\frac{\sqrt{n^2 + 5n}}{n}$ can be simplified to $\sqrt{1 + \frac{5}{n}}$. $\endgroup$ – Ted Feb 10 '20 at 23:38
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Using the binomial series, $$n\left(1+\dfrac5n\right)^{1/2}-n\left(1+\dfrac2n\right)^{1/2}=n\left(1+\dfrac12\dfrac5n+O\left[\dfrac1{n^2}\right]\right)-n\left(1+\dfrac12\dfrac2n+O\left[\dfrac1{n^2}\right]\right)\to\dfrac32.$$

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    $\begingroup$ Interesting approach, thanks for the answer! $\endgroup$ – Ted Feb 10 '20 at 23:43

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