Finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$

Question

I need help in finding the sum of the series $\sum_{n=2}^{\infty} ((n^2+1)^{1/2} - (n^3+1)^{1/3})$ if it converges.

I can't even prove convergence. I tried comparison test. I tried telescoping or even sandwich the partial sums. But I could not get anything.

I also tried using the identity

$a-b = \frac{a^3-b^3}{a^2+ab+b^2}$

and even the identity for $a^6-b^6$

but it won't yield.

I found this one while going through some really old question papers of my college.

Please help.

Answer 1

The point is "finding the sum if it converges".

If we are able to show that the sum doesn't converge, then there is no need to find its sum.

To show divergency, it suffices to use Taylor expansion to show that $(n^2 + 1)^{1/2} = n + \frac 1{2n} + O(\frac 1{n^2})$ and $(n^3 + 1)^{1/3} = n + O(\frac 1{n^2})$ when $n$ tends to infinity.

These together give $$(n^2 + 1)^{1/2} - (n^3 + 1)^{1/3} = \frac 1{2n} + O(\frac 1{n^2})$$ when $n$ tends to infinity. Thus the divergency follows from the fact that the harmonic series is divergent and the series $\sum\frac 1{n^2}$ is convergent.

Answer 2

Without series, we have

$$\sqrt{n^2+1}-\sqrt[3]{n^3+1} = \sqrt{n^2+1}- n + n - \sqrt[3]{n^3+1}$$

which allows us to use those conjugates without appealing to sixth order

$$= \frac{1}{n+\sqrt{n^2+1}} - \frac{1}{n^2+n\sqrt[3]{n^3+1}+(n^3+1)^{\frac{2}{3}}}$$

The divergence comes from the square root term.