Convergence for Infinite Series

[roam]

1. Examine the series $$\frac{1}{1 . 2} +$$$$\frac{1}{2 . 3}+$$$$\frac{1}{3 . 4}+$$$$\frac{1}{4 . 5}...$$ for convergence.



3. The Attempt at a Solution

The following is the book's answer:

"$$lim_{n\rightarrow \infty}S_{n}$$
$$lim_{n\rightarrow \infty} (1 - \frac{1}{n + 1}) = 1 - 0 = 1$$
Hence the series converge and its sum is 1. "


From $$S_{n} =$$$$\frac{1}{1 . 2} +$$$$\frac{1}{2 . 3}+$$$$\frac{1}{3 . 4}+$$$$\frac{1}{4 . 5}...$$

I can see that the nth term is $$\frac{1}{n . (n+1)}$$ but I can't follow how the book's obtained the "$$1 - \frac{1}{n + 1}$$" or th nth partial sum.

I appreciate some help. Unfortunently there are no explanations in the book on this question.

 

[rock.freak667]

try splitting $\frac{1}{n . (n+1)}$ into partial fractions, then use the property of a telescoping series.

 

[roam]

try splitting $\frac{1}{n . (n+1)}$ into partial fractions, then use the property of a telescoping series.

Is this correct:

$$(\frac{1}{1} \times \frac{1}{2}) + (\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{5}) + ... + (\frac{1}{n} \times \frac{1}{n + 1})$$

I still don't understand how to reach $$1 - \frac{1}{n + 1}$$

 

[NoMoreExams]

Did you rewrite 1/[n(n+1)] using partial fractions? Doesn't seem like you have. You just wrote out the first few terms and the n-th term.

 

[Дьявол]

$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

Now you got:

$$(1 - \frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n}-\frac{1}{n+1})$$

What can you assume?

 

[roam]

Oh, thanks. Now I understand what rockfreak was saying about splitting it into partial fractions.

$$(1 - \frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n}-\frac{1}{n+1})$$

What can you assume?

Now I can test for convergence in $$lim_{n \rightarrow \infty}(\frac{1}{n} - \frac{1}{n+a})$$ but the book says we need to have "$$1 - \frac{1}{n+1}$$" & I'm not sure where I can find it from.

 

[NoMoreExams]

No... look at what he wrote you see how for -1/2 there's a 1/2, for -1/3 there's a 1/3, etc. so what are you left with?

I believe the original post gave you 2 suggestions:
1) Use partial fractions.
2) Use properties of telescoping series.

1) was done for you by the Russian devil, now do part 2)

 

[roam]

Yes, they will cancel out and the series simply equal 1 therefore the series converge to one. I'll go & try some more examples & hopefully I think I learned the idea of "telescoping series"!

Thanks very much :)

 

[Дьявол]

Yes, they will cancel out and the series simply equal 1 therefore the series converge to one. I'll go & try some more examples & hopefully I think I learned the idea of "telescoping series"!

Thanks very much :)

No, look again.

There are $1 + \frac{1}{n+1}$ left.

Now you got $$\lim_{n \rightarrow \infty}(1 + \frac{1}{n+1})$$.

Because $$\lim_{n \rightarrow \infty}(\frac{1}{n+1})=0$$

you got what? (rhyme)

 

[sutupidmath]

$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$

Hold on here for a moment!



In the original post i can see that the series is actually

$$\sum_{n=1}^{\infty}\frac{1}{n.(n+1)}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...$$

and not

$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+...$$

There is a difference, don't u think? There certainly is a difference, unless the OP actually wanted to write the second one.

(To the OP:which one is it?)

 

[NoMoreExams]

He meant . as a a multiplication symbol, what he probably should have used was \cdot

 

[sutupidmath]

He meant . as a a multiplication symbol, what he probably should have used was \cdot

Well, yeah, that's what i was thinking too, since it isn't that easy to evaluate the sum of the other seires, unless we sought for some relation to a specific case of a Taylor expansion, or something like that.

 

[roam]

Hi StupidMath! Yes, like NomoreExams said I meant a multiplication symbol.


Sorry Dyavol, I forgot.

$$lim_{n\rightarrow \infty} (1 - \frac{1}{n+1}) = 1 - 0 = 1$$

Hence the series converges and its sum is 1.