- #1

icystrike

- 446

- 1

## Homework Statement

Qn 1:

[tex]\sum_{k=1}^{\infty}(-1)^{k}((\sqrt{k}+1)/(k+1))[/tex]

Interval of convergence for the series:

[tex]\sum_{k=0}^{\infty}((x^{k})/(\sqrt{(k²+3)}))[/tex]

The ans is [-1,1]

Last edited:

You should upgrade or use an alternative browser.

- Thread starter icystrike
- Start date

- #1

icystrike

- 446

- 1

Qn 1:

[tex]\sum_{k=1}^{\infty}(-1)^{k}((\sqrt{k}+1)/(k+1))[/tex]

Interval of convergence for the series:

[tex]\sum_{k=0}^{\infty}((x^{k})/(\sqrt{(k²+3)}))[/tex]

The ans is [-1,1]

Last edited:

- #2

Gib Z

Homework Helper

- 3,352

- 6

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

- #3

icystrike

- 446

- 1

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

Thanks Gib for the great hints.

The first one im 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 ive realised 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.

- #4

Gib Z

Homework Helper

- 3,352

- 6

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.

- #5

icystrike

- 446

- 1

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.

Uhh... Sorry Gib.. please help me on the first one agn.. I've made typo earlier.

(I hope i dont 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 [tex]\frac{1}{k}[/tex] which is a harmonic series.

Last edited:

- #6

Gib Z

Homework Helper

- 3,352

- 6

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.

- #7

icystrike

- 446

- 1

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 seperate sums ...

In fact i can say that [tex] \sum (-1)^k \frac{1}{k} > \sum (-1)^k \frac{1}{k+1} [/tex] (I think this is a wrong approach)

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

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

- #8

Gib Z

Homework Helper

- 3,352

- 6

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

- #9

icystrike

- 446

- 1

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

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

Share:

- Replies
- 5

- Views
- 431

- Last Post

- Replies
- 7

- Views
- 631

- Last Post

- Replies
- 4

- Views
- 491

- Replies
- 7

- Views
- 1K

- Replies
- 1

- Views
- 340

- Replies
- 21

- Views
- 1K

- Replies
- 17

- Views
- 827

- Last Post

- Replies
- 2

- Views
- 427

- Last Post

- Replies
- 4

- Views
- 520

- Replies
- 2

- Views
- 384