# Does this series converge? Using the limit comparison test

• Rectifier
In summary, the problem involves determining the convergence of a series by comparing it to a known convergent series using the limit comparison test. The attempt made so far involves simplifying the expression and finding the limit as k approaches infinity. The problem can also be solved using a telescoping series, but the goal is to use the limit comparison test.
Rectifier
Gold Member
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.

Rectifier said:
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}} ##.

Rectifier said:
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!

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##

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

Math_QED said:
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.

Rectifier
Mark44 said:
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.

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.

## What is the limit comparison test?

The limit comparison test is a method used to determine if an infinite series converges or diverges. It involves comparing the given series to a known series with known convergence properties.

## When is the limit comparison test used?

The limit comparison test is typically used when the series does not have easily identifiable terms or when other tests, such as the ratio or root test, are inconclusive.

## How does the limit comparison test work?

The limit comparison test works by taking the limit of the quotient of the given series and the known series. If the limit is a positive finite number, then the two series have the same convergence properties and will either both converge or both diverge.

## What is the known series used for comparison in the limit comparison test?

The known series used for comparison in the limit comparison test is typically a p-series, harmonic series, or geometric series. These are series with well-known convergence properties.

## What is the result of the limit comparison test?

The result of the limit comparison test is either a confirmation of convergence or divergence of the given series, or an indication that the test is inconclusive and another method should be used to determine convergence or divergence.

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