Check for absolute and conditional convergence $$\sum_{n=1}^{\infty }(-1)^n\left ( \sqrt{n^2+4n+1}-\sqrt{n^2+n+4} \right )$$

My attempt:

$$\sum_{n=1}^{\infty }(-1)^n\left ( \sqrt{n^2+4n+1}-\sqrt{n^2+n+4} \right )=$$ $$=\sum_{n=1}^{\infty }(-1)^n\left ( \sqrt{n^2+4n+1}-\sqrt{n^2+n+4} \right )\cdot \frac{\sqrt{n^2+4n+1}+\sqrt{n^2+n+4}}{\sqrt{n^2+4n+1}+\sqrt{n^2+n+4}}=$$ $$=\sum_{n=1}^{\infty }(-1)^n\frac{3n-3}{\sqrt{n^2+4n+1}+\sqrt{n^2+n+4}}$$

$$\frac{3n-3}{\sqrt{n^2+4n+1}+\sqrt{n^2+n+4}}=3\cdot \frac{n-1}{n\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+n\sqrt{1+\frac{1}{n}+\frac{4}{n^2}}}=\frac{3}{n}\cdot \frac{n-1}{n+2+\mathcal{O}\left ( \frac{1}{n} \right )+n+\frac{1}{2}+\mathcal{O}\left ( \frac{1}{n} \right )}=\frac{3}{n}\cdot \frac{n-1}{2n+\frac{3}{2}+\mathcal{O}\left ( \frac{1}{n} \right )}=\frac{3}{2}-\frac{27}{8n}+\mathcal{O}\left ( \frac{1}{n} \right )$$

I have found that the series does not converge and does not converge at all. Did I solve the problem correctly? Question: What can I say about conditional convergence?

  • $\begingroup$ A necessary condition for a series to converge is that its general term converges to $0$. You've proved that the general term does not converge to $0$. So yes you have proven that the series does not converge, even conditionally. $\endgroup$
    – Bruno B
    Jun 7, 2023 at 18:35

1 Answer 1


You're solution is correct.

You got $\sqrt{n^2+4n+1}-\sqrt{n^2+n+4}=\frac{3}{2}-\frac{27}{8n}+\mathcal{O}\left ( \frac{1}{n}\right)$. If we let $n\to\infty$ that means that general term approaches $\frac{3}{2}\neq0$. Which means that it doesn't only diverge absolutely also conditionally, because it doesn't satisfy one of the first conditions of convergence which is $\lim_{n\to\infty}{(\sqrt{n^2+4n+1}-\sqrt{n^2+n+4})}=0$


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