# Is the series covergent or divergent?

1. Sep 19, 2006

### shaiqbashir

is the series covergent or divergent???

I want to know that is the following series convergent or divergent??

$$\sum \frac{2}{\sqrt{n}+1}$$

when i apply divergent test to it, it comes equal to 0 , it means that divergent test gets failed. then how to solve it???

which test i should apply???

the correct answer in my copy is written as COnvergent, but im not getting a convergent answer.

2. Sep 19, 2006

### StatusX

Compare it to $\sum 1/n$.

3. Sep 19, 2006

### TD

Are you sure about that? Perhaps you should specify the limits, or may we assume from 0 (or 1?) to infinity?

4. Sep 19, 2006

### shaiqbashir

Thanks for ur Help StatusX!

I applied Limit comparison test and it works with series 1/n as Bn.

thanks a lot once again!!!

5. Sep 19, 2006

### TD

But you know that the harmonic series (1/n) diverges, right?

6. Sep 19, 2006

### shaiqbashir

okz okz! im sorry for some mistakes.

the correct answer of the above series is this that the series is DIVERGENT not convergent. it was my mistake!
So that means that Status X was correct, we can compare it with 1/n and we can also compare it with 1/(n^1/2)

both will give divergence as result by using Basic Comparison test.

Now im having another problem which looks like to me of the same sort. I dont know its correct answer. here is the problem :

$$\sum \frac{1}{\sqrt{n}+\sqrt{n+1}}$$

we have to find that is the following series converging or diverging??

here is what i have done:

The Divergence Test seems to be failed here as it is giving me an answer equals to zero!

Now if i go for basic comparison test, what series should i consider in order to compare it with above series???

I have worked on the following series and they dont give me proper results:

$$\sum \frac{1}{\sqrt{n}}$$

$$\sum \frac{1}{\sqrt{n+1}}$$

$$\sum \frac{1}{n}$$

$$\sum \frac{1}{n^(1.5)}$$

Then what should i consider???