Study the Convergence of $\sum \sqrt{n+a} - \sqrt[4]{n^2 + 2n}$

Question

First of all let $a \in R$

I tried using the root test and got nowhere. Also tried by comparisson but I get stuck. The last thing I tried is to rewrite the series as follows: $$ \sum (\sqrt{n+a} - \sqrt[4]{n^2 + 2n}) = \sum \frac{n+a - \sqrt{n^2 + 2n}}{\sqrt{n+a} + \sqrt[4]{n^2 + 2n}} $$ since I've seen lots of problems similar to this being treated this way. I assume that I need to compare it to a $p$-series. I need a hint.

Answer 1

I would use the fact that $x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$. So:\begin{multline}\sqrt{n+a}-\sqrt[4]{n^2+2n}=\sqrt[4]{n^2+2an+a^2}-\sqrt[4]{n^2+2n}=\\=\frac{2an-2n+a^2}{\sqrt[4]{n^2+2an+a^2}^3+\sqrt[4]{n^2+2an+a^2}^2\sqrt[4]{n^2+2n}+\sqrt[4]{n^2+2an+a^2}\sqrt[4]{n^2+2n}^2+\sqrt[4]{n^2+2n}^3}\end{multline}Since the denominator behaves as $4n^{\frac32}$, the series converges if $a=1$ and diverges otherwise.

Answer 2

Note that by binomial expansion

• $\sqrt{n+a}=\sqrt n(1+a/n)^\frac12=\sqrt n+\frac{a}{2\sqrt n}-\frac{a^2}{8n\sqrt n}+o\left(\frac{1}{n\sqrt n}\right)$

• $\sqrt[4]{n^2+2n}=\sqrt n(1+2/n)^\frac14=\sqrt n+\frac1{2\sqrt n}-\frac{3a^2}{8n\sqrt n}+o\left(\frac{1}{n\sqrt n}\right)$

then

$$\sqrt{n+a} - \sqrt[4]{n^2 + 2n}=\frac1{2\sqrt n}\left(a-1\right)+\frac{a^2}{4n\sqrt n}+o\left(\frac{1}{n\sqrt n}\right)$$

and for $a\neq 1$ the series diverges and converges for $a=1$.

Answer 3

Go on: \begin{align} \sum (\sqrt{n+a} - \sqrt[4]{n^2 + 2n}) &= \sum \frac{n+a - \sqrt{n^2 + 2n}}{\sqrt{n+a} + \sqrt[4]{n^2 + 2n}} \\[6px] &=\sum\frac{n^2+2an+a^2-n^2-2n}{(\sqrt{n+a} + \sqrt[4]{n^2 + 2n})(n+a + \sqrt{n^2 + 2n})} \\[6px] &=\sum \frac{2(a-1)n+a^2}{ n\sqrt{n}(\sqrt{1+a/n}+\sqrt[4]{1+2/n})(1+a/n+\sqrt{1+2/n}) } \end{align}