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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?

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3 Answers 3

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

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

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  • $\begingroup$ How did you know that last step? why did you choose the exponent to $3/2$? $\endgroup$
    – Parseval
    Commented Dec 31, 2017 at 17:06
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    $\begingroup$ @Parseval Use $\sqrt{k + \frac1k} \ge \sqrt{k}$. $$\sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k+\frac{1}{k}}+\sqrt{k}} \le \sum_{k=1}^\infty \frac{\frac1k}{\sqrt{k}+\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k\sqrt{k}} = \sum_{k=1}^\infty \frac{1}{2k^{3/2}}$$ $\endgroup$ Commented Dec 31, 2017 at 17:09
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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.

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  • $\begingroup$ I don't see what you did when you introduced the ordo. How did you get rid of the square root? $\endgroup$
    – Parseval
    Commented Dec 31, 2017 at 17:00
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    $\begingroup$ I’ve used the binomial expansion $(1+x)^n=1+nx+o(x)$ $\endgroup$
    – user
    Commented Dec 31, 2017 at 17:13
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    $\begingroup$ in this way you can see the behavior of the tail of the series for k large which governs the convergence $\endgroup$
    – user
    Commented Dec 31, 2017 at 17:17
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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 $1.5>1$.

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  • $\begingroup$ 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? $\endgroup$
    – Parseval
    Commented Dec 31, 2017 at 16:57
  • $\begingroup$ Just $k\cdot\sqrt{k}=k^{1.5}$ and $\sum\limits_{k=1}^{\infty}\frac{1}{k^{1.5}}$ converges. $\endgroup$ Commented Dec 31, 2017 at 17:29
  • $\begingroup$ But it's also $k\cdot \sqrt{k+1/k}.$ $\endgroup$
    – Parseval
    Commented Dec 31, 2017 at 17:40
  • $\begingroup$ It's not relevant because $k+\frac{1}{k}>k$ and $\frac{1}{k\left(\sqrt{k+\frac{1}{k}}+\sqrt{k}\right)}<\frac{1}{k^{1.5}}$. $\endgroup$ Commented Dec 31, 2017 at 17:42

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