# 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 $3/2$? Dec 31, 2017 at 17:06
• @Parseval Use $\sqrt{k + \frac1k} \ge \sqrt{k}$. $$\sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k+\frac{1}{k}}+\sqrt{k}} \le \sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k}+\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k^{3/2}}$$ Dec 31, 2017 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? Dec 31, 2017 at 17:00
• I’ve used the binomial expansion $(1+x)^n=1+nx+o(x)$
– user
Dec 31, 2017 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, 2017 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? Dec 31, 2017 at 16:57
• Just $k\cdot\sqrt{k}=k^{1.5}$ and $\sum\limits_{k=1}^{\infty}\frac{1}{k^{1.5}}$ converges. Dec 31, 2017 at 17:29
• But it's also $k\cdot \sqrt{k+1/k}.$ Dec 31, 2017 at 17:40
• It's not relevant because $k+\frac{1}{k}>k$ and $\frac{1}{k\left(\sqrt{k+\frac{1}{k}}+\sqrt{k}\right)}<\frac{1}{k^{1.5}}$. Dec 31, 2017 at 17:42