Does this series converge? Using the limit comparison test

[Rectifier]

The problem
In this problem I am supposed to show that the following series converges by somehow comparing it to $\frac{1}{k\sqrt{k}}$ :
$$ \sum^{\infty}_{k=1} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \right) $$

The attempt
$\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}$

Now, I am supposed to somehow find:
$\lim_{k \rightarrow \infty} \frac{\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}}{\frac{1}{k\sqrt{k}}}$. I am not sure that this I have simplified the starting expression enough.

There are alternative solutions to this problem but I would like to solve it using limit comparison test. Thank you for your understanding.

 

[Dick]

The problem
In this problem I am supposed to show that the following series converges by somehow comparing it to $\frac{1}{k\sqrt{k}}$ :
$$ \sum^{\infty}_{k=1} \left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \right) $$

The attempt
$\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} = \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}$

Now, I am supposed to somehow find:
$\lim_{k \rightarrow \infty} \frac{\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}}{\frac{1}{k\sqrt{k}}}$. I am not sure that this I have simplified the starting expression enough.

There are alternative solutions to this problem but I would like to solve it using limit comparison test. Thank you for your understanding.

You're right. You haven't simplified it enough. Try multiplying numerator and denominator of the attempt by $\sqrt{k+1}+\sqrt{k}$.

 

[Rectifier]

I get:
$\frac{1}{(k+1)\sqrt{k+1} + k\sqrt{k+1}}$.

Can I now assume that the expression behaves as follows for big k?
$\frac{1}{(k+1)\sqrt{k+1} + k\sqrt{k+1}} \approx \frac{1}{2k\sqrt{k}}$.

 

[Dick]

I get:
$\frac{1}{(k+1)\sqrt{k+1} + k\sqrt{k+1}}$.

Can I now assume that the expression behaves as follows for big k?
$\frac{1}{(k+1)\sqrt{k+1} + k\sqrt{k+1}} \approx \frac{1}{2k\sqrt{k}}$.

It clearly does behave that way. Whether you can 'assume' it or not depends on what your instructor expects.

 

[Rectifier]

I guess that I will show that the first denomenator is bigger than the first first by using the first comparison test. I could then show that it is the bigger one and converges then so does the smaller one. I could now te limit somparison test on the bigger one to show that it converges.

Thank you so much for your help!

 

[member 587159]

Hold on! If you see something where terms with $k-1$ and $k$ are canceled out consecutively, a red flag must appear in your brain! Stop thinking about it, you can calculate this sum!

Denote $S_n$ with the $n$-th partial sum. Then

$S_n = \sum_{k=1}^n \left(\frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k-1}}\right)= \frac{1}{\sqrt{1}}- \frac{1}{\sqrt{n+1}} \to 1$

So $\sum_{k=1}^\infty \left(\frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k-1}}\right) = 1$

 

[Mark44]

In this problem I am supposed to show that the following series converges by somehow comparing it to $\frac{1}{k\sqrt{k}}$

Hold on! If you see something where terms with k−1 and k are canceled out consecutively, a red flag must appear in your brain! Stop thinking about it, you can calculate this sum!

From the problem statement in post #1, the OP doesn't need to calculate the sum--just show that the series converges. Also, the technique to use, the limit comparison test, is specified.
I agree that this is a "telescoping series" that converges, but that's not relevant in this problem, according to the problem statement.

 

[member 587159]

From the problem statement in post #1, the OP doesn't need to calculate the sum--just show that the series converges. Also, the technique to use, the limit comparison test, is specified.
I agree that this is a "telescoping series" that converges, but that's not relevant in this problem, according to the problem statement.

You are right. Nevertheless, it might be interesting to the OP, to already have seen this trick.

 

[Rectifier]

Thank you too everyone. I was looking for that specific answer that Dick helped me with. The one Math QED provided was an exercise a a bit earlier in my book, which I managed to solve. I was hoping that the diclaimer I provided would help homework helpers to not waste their time.Thank you for your effort nevertheless.

There are alternative solutions to this problem but I would like to solve it using limit comparison test.