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We finished covering several tests that help determine the convergence or divergence of a series and I tried them all and I couldn't make progress or produce an answer that I felt was coherent enough to follow.

there are two series up for consideration, the first;

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

and

$\sum_{k=1}^{\infty} ln(1+\frac{1}{2^k})$

For the first one I am not sure what test would work best but on the second one because there is a number number raised to the power $k$ I am led to believe the root test would work well, but I struggled to use algebra to work around the $ln$

Any hints or tips for proceeding would be highly appreciated.

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    $\begingroup$ Use the comparison test, and don't forget that $\dfrac{\sqrt{k+1} + \sqrt{k}}{\sqrt{k+1} + \sqrt{k}} = 1$. $\endgroup$ – Daniel Fischer Sep 6 '16 at 13:07
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For the first series, \begin{align*} \sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt k} &=\sum_{k=1}^\infty\frac{(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)}{\sqrt k(\sqrt{k+1}+\sqrt k)}\\ &=\sum_{k=1}^\infty\frac1{\sqrt k(\sqrt{k+1}+\sqrt k)}\\ &\ge\sum_{k=1}^\infty\frac1{\sqrt k(\sqrt{k+k}+\sqrt k)}\\ &=\frac1{1+\sqrt 2}\sum_{k=1}^\infty\frac1k=\infty. \end{align*} For the second series, we can use the inequality $$ \log(1+x)<x $$ for $x>0$. We obtain $$ \sum_{k=1}^\infty\log\biggl(1+\frac1{2^k}\biggr)<\sum_{k=1}^\infty\frac1{2^k}=1. $$

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Hints:

First series:

$$\frac{\sqrt{k+1}-\sqrt k}{\sqrt k}=\frac{1}{\sqrt{k^2+k}+k}>\frac1{2k+1}$$

Second series:

Cauchy condensation test implies that this series converges iff the following either does:

$$\sum_{n=1}^\infty\frac{\log(1+\frac1n)}{2^n}$$

And for $t>0$, $$\log(1+t)=\int_1^{1+t}\frac{du}u<t$$

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