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I have this series: $$\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$$ I know that it's convergent (from WolframAlpha) but I need to prove it is convergent. How can I do it?

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You have that $$\begin{align*} \dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}&=\frac1n+\frac{(\sqrt{1+n^2}-\sqrt{2+n^2})\cdot(\sqrt{1+n^2}+\sqrt{2+n^2})}{\sqrt{1+n^2}+\sqrt{2+n^2}}=\\\\&=\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\\& \le \frac{1}{n} \end{align*}$$ Then the convergence is established by squaring both sides and using the comparison test.


For the above implication to be correct we also need to show that the middle part is positive. Indeed $$\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ < \frac{1}{\sqrt{n^2}+\sqrt{n^2}}=\frac{1}{2n}$$ or equivalently $$-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\ >-\frac{1}{2n}$$ which implies that $$\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}>\frac1n-\frac{1}{2n}=\frac1{2n}>0$$

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  • $\begingroup$ But comparison test needs $\sum_1^n\frac1n$ to be convergent, but it isn't. @Stefanos $\endgroup$ – michaeluskov Apr 8 '14 at 12:08
  • $\begingroup$ @user23791 Square both sides firstly.... $\endgroup$ – Jimmy R. Apr 8 '14 at 12:08
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    $\begingroup$ Simple and elegant, but you have to show that middle part is positive, not just smaller than $\frac{1}{n}$ (It is too obvious, but maybe for the OP) $\endgroup$ – derivative Apr 8 '14 at 12:14
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Hint: $\dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}\sim _\infty \dfrac 1 n$.

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  • $\begingroup$ But $\sum_{n=1}^{\infty}\frac1n$ doesn't converge. How will it help me? $\endgroup$ – michaeluskov Apr 8 '14 at 11:27
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    $\begingroup$ @user23791 Don't forget the square. $\endgroup$ – Git Gud Apr 8 '14 at 11:27
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    $\begingroup$ @user23791 If you wish you can read my hint as $\left(\dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2\sim _\infty \left(\dfrac 1 n\right)^2$. $\endgroup$ – Git Gud Apr 8 '14 at 12:09

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