# Does the series $\sum_{k=1}^{\infty}\left(\sqrt{k+\frac{1}{k}}-\sqrt{k}\right)$ converge or diverge?

Does the series $\sum_{k=1}^{\infty}\left(\sqrt{k+\frac{1}{k}}-\sqrt{k}\right)$ converge or diverge?

What test would be the most appropriate for this series? I've tried the ratio, root and integral tests but no luck. Would Mclaurin expansion work here?

It converges. Just multiply each term with $\frac{\sqrt{k+\frac{1}{k}}+\sqrt{k}}{\sqrt{k+\frac{1}{k}}+\sqrt{k}}$.

$$\sum_{k=1}^{\infty}\left(\sqrt{k+\frac{1}{k}}-\sqrt{k}\right) = \sum_{k=1}^\infty \frac{k+ \frac{1}{k} - k}{\sqrt{k+\frac{1}{k}}+\sqrt{k}} = \sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k+\frac{1}{k}}+\sqrt{k}} \le \sum_{k=1}^\infty \frac{1}{2k^{3/2}} < +\infty$$

• How did you know that last step? why did you choose the exponent to $3/2$? – Parseval Dec 31 '17 at 17:06
• @Parseval Use $\sqrt{k + \frac1k} \ge \sqrt{k}$. $$\sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k+\frac{1}{k}}+\sqrt{k}} \le \sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k}+\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k^{3/2}}$$ – mechanodroid Dec 31 '17 at 17:09

Note that

$$\left(\sqrt{k+\frac{1}{k}}-\sqrt{k}\right)=\sqrt{k}\left(\sqrt{1+\frac{1}{k^2}}-1\right)=\sqrt{k}\left(1+\frac{1}{2k^2}+o\left(\frac{1}{k^2}\right)-1\right)=\\=\frac{1}{2k^{\frac32}}+o\left(\frac{1}{k^{\frac32}}\right)$$

Thus, the given series converges.

• I don't see what you did when you introduced the ordo. How did you get rid of the square root? – Parseval Dec 31 '17 at 17:00
• I’ve used the binomial expansion $(1+x)^n=1+nx+o(x)$ – user Dec 31 '17 at 17:13
• in this way you can see the behavior of the tail of the series for k large which governs the convergence – user Dec 31 '17 at 17:17

It converges of course because $$\sqrt{k+\frac{1}{k}}-\sqrt{k}=\frac{1}{k\left(\sqrt{k+\frac{1}{k}}+\sqrt{k}\right)}$$ and $1.5>1$.

• I did the same as you did, multplying by the conjugate but I never realised that the exponent sums up to 1.5 in the numerator, how do you know it's 1.5? – Parseval Dec 31 '17 at 16:57
• Just $k\cdot\sqrt{k}=k^{1.5}$ and $\sum\limits_{k=1}^{\infty}\frac{1}{k^{1.5}}$ converges. – Michael Rozenberg Dec 31 '17 at 17:29
• But it's also $k\cdot \sqrt{k+1/k}.$ – Parseval Dec 31 '17 at 17:40
• It's not relevant because $k+\frac{1}{k}>k$ and $\frac{1}{k\left(\sqrt{k+\frac{1}{k}}+\sqrt{k}\right)}<\frac{1}{k^{1.5}}$. – Michael Rozenberg Dec 31 '17 at 17:42