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;

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


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

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.

  • 2
    $\begingroup$ 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

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 $$ \sum_{k=1}^\infty\log\biggl(1+\frac1{2^k}\biggr)<\sum_{k=1}^\infty\frac1{2^k}=1. $$



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:


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.