# Proving the series $\sum \limits_{n=1}^{\infty}\left( \sqrt{1+n^{4}}-n^{2}\right)$ is convergent.

I need to prove that the series $\sum \limits_{n=1}^{\infty}\left( \sqrt{1+n^{4}}-n^{2}\right)$ is convergent. But I have kown idea how to prove it. Please anyone give me a help. Thanks.

• Just to get your thoughts out, may I ask why you think it is convergent? – Sujaan Kunalan Mar 11 '15 at 15:54
• If you have no idea how to prove it, why do you think it's convergent...? – Hans Lundmark Mar 11 '15 at 16:20

Just write

$$0\le\sqrt{1+n^{4}}-n^{2}=\frac{1}{\sqrt{1+n^{4}}+n^{2}}\le \frac1{2n^2}$$ and use the comparison with the convergent Riemann series $\sum\frac1{n^2}$.

Hint: Observe that for each $n\in \mathbb{N}$,

$$\left| \sqrt{1+n^{4}}-n^{2}\right|=\sqrt{1+n^{4}}-n^{2}=\dfrac{(\sqrt{1+n^{4}}-n^{2})(\sqrt{1+n^{4}}+n^{2})}{\sqrt{1+n^{4}}+n^{2}}=\dfrac{1}{\sqrt{1+n^{4}}+n^{2}}<\dfrac{1}{n^{2}}$$

and

$$\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}} \text{ is convergent.}$$

Let $$S = \sum_{n=1}^\infty \left( \sqrt{1+n^4}-n^2\right) = \sum_{n=1}^\infty s_n$$ For all $n>0$, $$1+n^4 > n^4 = (n^2)^2 \Rightarrow \sqrt{1+n^4}-n^2 >0$$ so $S$ is the sum of positive terms, thus to show $S$ is convergent we must merely demonstrate a positive upper bound on $S$. We will do this by demonstrating that the $s_m$ are all less than $t_n$ for a series $T$ which we know to converge:

$$\sqrt{1+n^4} = n^2 \sqrt{1+\frac{1}{n^4} }< n^2 \left( 1 + \frac{1}{2n} \right) = n^2 + \frac{1}{2n}$$ $$s_n < \frac{1}{2n^2}$$ and we know that the sum of $\frac{1}{n^2}$ is finite.