convergence of $ \sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} } $

Question

I want to prove that $\sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} }$ converges. How can I do that?

I have simplified that into:

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

If I show that $\sum_{k=1}^{\infty}{ \frac{1}{k^2+k} }$ converges then $\sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} }$ must converge, right?

How can I go on ?

I have tried the quotient criterium which gives me equals $1$. So it is not very helpful here.

Answer 1

$$\sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} } > \sum_{k=1}^{\infty}{ \frac{1}{\sqrt{(k+1)(k+1)}}} = \sum_{k=1}^{\infty}{ \frac{1}{k+1} }$$

$$\implies \sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} } > \sum_{k=1}^{\infty}{ \frac{1}{k+1} }$$

Since $\displaystyle \sum_{k=1}^{\infty}{ \frac{1}{k+1} }$ diverges, $\displaystyle \sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} }$ diverges too.

Answer 2

$$\frac{1}{\sqrt{k(k+1)}} = \frac{1}{k}\cdot \frac{1}{\sqrt{1+\frac{1}{k}}} > \frac{1}{k}$$

meaning you will have fairly little luck in proving convergence...

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Also, you are very very wrong in saying

If I show that $\sum_{k=1}^{\infty}{ \frac{1}{k^2+k} }$ converges then $\sum_{k=1}^{\infty}{ \frac{1}{\sqrt{k(k+1)}} }$ must converge, right?

The answer to that question is NO. Similarly, you can show that $$\sum_{k=1}^\infty\frac{1}{k^2}$$ converges, but $$\sum_{k=1}^\infty\frac1k$$does not.

Answer 3

$a_k:= \dfrac{1}{\sqrt{(k+1)^2}} \lt \dfrac{1}{\sqrt{k(k+1)}}.$

$\sum_{k=1}^{\infty} a_k$ is divergent , harmonic series.

Answer 4

With equvalents:

$\sqrt{k(k+1)}\sim_\infty\sqrt{k^2}=k$, so $\;\dfrac 1{\sqrt{k(k+1)}}\sim_\infty\frac 1k,\; $ and the harmonic series diverges.