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I'm trying to determine if the following series converges or diverges.

$$\sum_{k=1}^\infty\frac{\sqrt{1+k}}{k}-\frac{1}{\sqrt{k}}$$

Now, I've tried the obvious things, such as the nth term test, trying to write the expression into a single fraction, then apply a comparison test with $\frac{1}{k}$ to get rid of the denominator, however upon taking the limit, it then yeilds zero, which means the test is inconclusive. Furthermore, I don't think the expression is easily integrable, nor can we apply some Taylor or Maclaurin series.

Ideas?

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3 Answers 3

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$\frac {\sqrt {1+k}} k-\frac 1 {\sqrt k}=\frac {\sqrt {k^{2}+k} -k} {k\sqrt k}=\frac {\sqrt {1+\frac1 k} -1} {\sqrt k}$. Now observe that $\sqrt {1+\frac1 k} <1+\frac 1 {k}$. This gives $0 \leq \frac {\sqrt {1+k}} k-\frac 1 {\sqrt k} <\frac 1 {k\sqrt k}$ hence, the series is convergent.

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  • $\begingroup$ Bernoulli's Inequality says $\sqrt{1+\frac1k}\le\left(1+\frac1{2k}\right)$ giving an even tighter bound. $\endgroup$
    – robjohn ♦
    Oct 4 at 13:09
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Observing $$ 0<\frac {\sqrt {1+k}} k-\frac 1 {\sqrt k}=\frac {\sqrt {k^{2}+k} -k} {k\sqrt k}=\frac {\sqrt {1+\frac1 k} -1} {\sqrt k}=\frac 1{(\sqrt {1+\frac1 k} +1)k^{3/2}}<\frac1{k^{3/2}}, $$ one concludes that the series converses.

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$\textbf{Hint:}$ We have the inequality

$$0 \leq \frac{\sqrt{1+k}}{k} - \frac{1}{\sqrt{k}} \leq \frac{1}{k\sqrt{k}}$$

For all $k \geq 1$

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