Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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 $$\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}>\frac1n-\frac{1}{2n}=\frac1{2n}>0$$

share|improve this answer
    
But comparison test needs $\sum_1^n\frac1n$ to be convergent, but it isn't. @Stefanos –  michaeluskov Apr 8 at 12:08
    
@user23791 Square both sides firstly.... –  Stef Apr 8 at 12:08
2  
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) –  derivative Apr 8 at 12:14

Hint: $\dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}\sim _\infty \dfrac 1 n$.

share|improve this answer
    
But $\sum_{n=1}^{\infty}\frac1n$ doesn't converge. How will it help me? –  michaeluskov Apr 8 at 11:27
3  
@user23791 Don't forget the square. –  Git Gud Apr 8 at 11:27
1  
@user23791 If you wish you can read my hint as $\left(\dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2\sim _\infty \left(\dfrac 1 n\right)^2$. –  Git Gud Apr 8 at 12:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.