Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem.
Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$ But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
Best Answer
You cannot prove $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1 $$ because the left-hand side tends to $+\infty$ for $n \to \infty$.
You should compare your series with $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ instead: $$ \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{2}{n^{3/2}} \cdot \frac{1}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} <\frac{2}{n^{3/2}} \, . $$