Does the series $\sum_{n=1}^{\infty} (n^2+1)^{1/2}-(n^3+1)^{1/3}$ converge?
I am not able to use the standard tests on this series. Any hints/advice will be much appreciated.
The solution hint is:
I don't understand how one get's this expression. The first fraction is clear but the last fraction is not so clear. What did the author use to rationalize the expression?
On
Use development of in Taylor Series, you will have your result. For example $$ \sqrt[3]{1+n^3}=n\sqrt[3]{1+\frac{1}{n^3}}=n\left(1+\frac{1}{3n^3}+o\left(\frac{1}{n^3}\right)\right) $$
Do the same for the square root and tell us if it converges or not. Identically,
$$ \sqrt{1+n^2}=n\sqrt[2]{1+\frac{1}{n^2}}=n\left(1+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right) $$ Hence $$ \sqrt{1+n^2}-\sqrt[3]{1+n^3}\underset{(+\infty)}{=}n+\frac{1}{2n}-n-\frac{1}{3n^2}+o\left(\frac{1}{n}\right) $$ Hence sadly
$$ \sqrt{1+n^2}-\sqrt[3]{1+n^3}\underset{(+\infty)}{\sim}\frac{1}{2n}$$
The series $\displaystyle \sum_{ \geq 1}^{ }\frac{1}{n}$ diverges so as your series.
On
I take this chance as a teaching opportunity.
Convergence tests are just condensed forms of chains of inequalities. As a thumb rule, if you cannot figure out which convergence test is best suited for the task at hand, it is worth to take a step back and just try to tackle your problem by elementary inequalities. First consideration: for any $n\geq 1$, both $\sqrt{n^2+1}$ and $\sqrt[3]{n^3+1}$ are pretty close to $n$, so it looks reasonable that the convergence of the given series depends on how much they are close to $n$. We have $$\sqrt{n^2+1}-n = \frac{1}{n+\sqrt{n^2+1}}\approx \frac{1}{2n} $$ $$\sqrt[3]{n^3+1}-n = \frac{1}{n^2+n\sqrt[3]{n^3+1}+\sqrt[3]{n^3+1}^2}\approx \frac{1}{3n^2} $$ due to $(a^2-b^2)=(a-b)(a+b)$ and $(a^3-b^3)=(a-b)(a^2+ab+b^2)$. The series $\sum_{n\geq 1}\frac{1}{3n^2}$ is convergent, but the series $\sum_{n\geq 1}\frac{1}{n}$ is not, so the given series is expected to be divergent. In order to turn this second consideration into a rigorous proof, it is enough to replace the first approximation above with $$\forall n\geq 1,\qquad \sqrt{n^2+1}-n \geq \frac{1}{3n} $$ implying that $\sum_{n\geq 1}\sqrt{n^2+1}-\sqrt[3]{n^3+1}$ is really divergent.
Since 'the first fraction is clear', the question about the hint seems to be if the equality below is true, $$ n^2+1 - \sqrt[3]{(n^3+1)^2}=\frac{3n^4 - 2n^3+3n^2}{(n^2+1)^2+(n^2+1)\sqrt[3]{(n^3+1)^2} + (n^3+1)\sqrt{n^3+1}}$$ Observe that if we define $a=\sqrt{n^2+1}>0\, ,b=\sqrt[3]{n^3+1}>0$, this is asking if $$ a^2 - b^2 = \frac{3n^4 - 2n^3 + 3n^2}{a^4 + a^2 b^2 + b^4} $$ which is equivalent to $$ (a^2-b^2)(a^4 + a^2b^2 + b^4) =3n^4 - 2n^3 + 3n^2$$ we rewrite the left hand side, \begin{align} LHS &= (a^2-b^2)(a^4 + a^2b^2 + b^4) \\&= a^6 + \color{red}{a^4b^2}+\color{blue}{a^2b^4}-\color{red}{a^4b^2}-\color{blue}{a^2b^4}-b^6\\ &= a^6 - b^6\\ &=(n^2+1)^3 - (n^3+1)^2\\ &= (n^6+3n^4+3n^2+1) - (n^6 + 2n^3 + 1)\\ &= 3n^4 - 2n^3 + 3n^2\\ &= RHS \end{align} as needed. I much prefer the other answers, but there you go.