# Converges or Diverges? Relavant test

• icystrike
In summary, the first series is an alternating series and can be proven to be convergent using the Leibniz criterion. For the second series, the ratio test can be used to show that the interval of convergence is (-1,1), with the endpoints -1 and 1 being dealt with by the Leibniz criterion and comparison to a divergent series. The lecturer's given answer of [-1,1] may be incorrect. The Leibniz criterion can be applied directly without the need for comparison. The ratio of consecutive terms for the second series can be shown to be less than 1, indicating convergence.

## Homework Statement

Qn 1:
$$\sum_{k=1}^{\infty}(-1)^{k}((\sqrt{k}+1)/(k+1))$$

Interval of convergence for the series:
$$\sum_{k=0}^{\infty}((x^{k})/(\sqrt{(k²+3)}))$$
The ans is [-1,1]

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For the first one - it is an alternating series. Have you tried to apply the Leibniz criterion?

For the second one - use the ratio test. And the answer isn't [-1,1].

Gib Z said:
For the first one - it is an alternating series. Have you tried to apply the Leibniz criterion?

For the second one - use the ratio test. And the answer isn't [-1,1].

Thanks Gib for the great hints.

The first one I am not able to prove that the series is monotone decreasing. (I've tried using triangle inequality). Will take a look again.. meanwhile do u mind sheding more light on this? (Btw I've realized that I've made some typo error please look at the series agn)

The second one that is the answer given by the lecturer. I've gotten [-1,1) instead.

If I am reading correctly, is the term in your first one $$\frac{ \sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k+1}}$$ ? If so, it should be evident that it is decreasing, but you can also check the difference of consecutive terms is negative quite easily.

For the second one, you have the right answer. The inside interval (-1,1) follows from the ratio test, the endpoint -1 follows from the Leibniz criterion, and the endpoint 1 is dealt with by comparison to a well known divergent series. Your lecturer probably made a silly mistake, and will be impressed if you also add in the proof of why it is divergent at x=1.

Gib Z said:
If I am reading correctly, is the term in your first one $$\frac{ \sqrt{k+1}}{k+1} = \frac{1}{\sqrt{k+1}}$$ ? If so, it should be evident that it is decreasing, but you can also check the difference of consecutive terms is negative quite easily.

For the second one, you have the right answer. The inside interval (-1,1) follows from the ratio test, the endpoint -1 follows from the Leibniz criterion, and the endpoint 1 is dealt with by comparison to a well known divergent series. Your lecturer probably made a silly mistake, and will be impressed if you also add in the proof of why it is divergent at x=1.

(I hope i don't not have to resort to second finte induction (Mathematical induction)

For the second one, I've proved that at x=1 the series is divergent by limit comparision test with $$\frac{1}{k}$$ which is a harmonic series.

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You can see that $$\sum (-1)^k \frac{1}{k+1}$$ converges, so we need only check $$\sum (-1)^k \frac{ \sqrt{k}}{k+1}$$ converges or diverges, since by addition we get the original series back (of course, we use "Convergent + Convergent = Convergent, Convergent + Divergent = Divergent").

So for the simpler one we are checking, by the Leibniz criterion we must check if the term is eventually decreasing. One way to do this is to show the ratio of consecutive terms is less than 1, so try to show that.

Gib Z said:
You can see that $$\sum (-1)^k \frac{1}{k+1}$$ converges, so we need only check $$\sum (-1)^k \frac{ \sqrt{k}}{k+1}$$ converges or diverges, since by addition we get the original series back (of course, we use "Convergent + Convergent = Convergent, Convergent + Divergent = Divergent").

So for the simpler one we are checking, by the Leibniz criterion we must check if the term is eventually decreasing. One way to do this is to show the ratio of consecutive terms is less than 1, so try to show that.

Oh! thanks! I was reluctant i split into separate sums ...
In fact i can say that $$\sum (-1)^k \frac{1}{k} > \sum (-1)^k \frac{1}{k+1}$$ (I think this is a wrong approach)

Then i can prove that $$\sum (-1)^k \frac{1}{k}$$ is convergent by leibniz criterion..

However my question is whether the comparision test is only suitable for positive terms?

Don't do comparison, just apply Leibniz straight away. 1/(k+1) is decreasing.

Show the ratio of consecutive terms is less than 1 for $$\sum (-1)^k \frac{\sqrt{k}}{k+1}$$ too.

Gib Z said:
Don't do comparison, just apply Leibniz straight away. 1/(k+1) is decreasing.

Show the ratio of consecutive terms is less than 1 for $$\sum (-1)^k \frac{\sqrt{k}}{k+1}$$ too.

But we can only show the ratio is less than 1 for finite terms is it sufficient?

## 1. What does it mean for a series to converge or diverge?

Convergence and divergence refer to the behavior of a mathematical series, which is a sequence of numbers that are added together. If the terms of the series approach a finite number as the number of terms increases, the series is said to converge. If the terms of the series do not approach a finite number, the series is said to diverge.

## 2. How do you determine if a series converges or diverges?

To determine if a series converges or diverges, various tests can be used, such as the comparison test, the integral test, or the ratio test. These tests involve evaluating the behavior of the series and comparing it to known patterns of convergence or divergence.

## 3. What is the relevant test for determining if a series converges or diverges?

The relevant test for determining if a series converges or diverges depends on the specific characteristics of the series. For example, if the terms of the series decrease in size and approach 0, the series can be tested using the integral or comparison tests. If the terms of the series alternate between positive and negative values, the alternating series test can be used.

## 4. How do you use the relevant test to determine convergence or divergence?

To use the relevant test for determining convergence or divergence, the behavior of the series must be examined and compared to the criteria of the test. If the criteria are met, the series can be classified as either convergent or divergent. If the criteria are not met, the test may not be applicable and another test should be considered.

## 5. Can a series both converge and diverge?

No, a series cannot both converge and diverge. It can only exhibit one of these behaviors. However, some series may not exhibit either behavior and are considered to be indeterminate. In these cases, more advanced techniques may be needed to determine the behavior of the series.