# Thread: Analize convergence of an infinite series

1. ## Analize convergence of an infinite series

This is the series:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}$

So this is what I did:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= \tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}$
So the first term diverges because it is harmonic and thus the hole series diverges.

I have this marked as wrong, but I don't understand why. Can somebody explain it please?

2. Originally Posted by ascii
This is the series:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}$

So this is what I did:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= \tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}$
So the first term converges because it is harmonic and thus the hole series converges.

I have this marked as wrong, but I don't understand why. Can somebody explain it please?
Because if you actually write down the first couple terms of the series, you will see all but the first and the last terms cancel out

Oh, and FYI $\displaystyle \infty$ is given by \infty =)

3. Oh thank you... so if there was a plus between the series they would diverge for sure.
I'll try other way.

Thanks again

4. So the first term converges because it is harmonic
If you're talking about the harmonic series, then it diverges! (you can see it by doing the integral test for example).

5. Originally Posted by arbolis
If you're talking about the harmonic series, then it diverges! (you can see it by doing the integral test for example).
Oopsy! I know that, my mistake typing. I meant diverges.
If I had meant converges, then the question in the first place wouldn't have made any sense.

6. Ok so I got this series:
$\displaystyle \tfrac{1}{2}\sum_{n=1}^{\infty}{ (\tfrac{1}{n(n+1)})}$
Which converges as the limit comparison with 1/n^2 confirms.
But now I'm asked to express the result of the sum.
I think this is a telescoping series, isn't it?
If it is, then the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?

7. Originally Posted by ascii
Ok so I got this series:
$\displaystyle \tfrac{1}{2}\sum_{n=1}^{\infty}{ (\tfrac{1}{n(n+1)})}$
Which converges as the limit comparison with 1/n^2 confirms.
But now I'm asked to express the result of the sum.
I think this is a telescoping series, isn't it?
If it is, then the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} = 1 - \frac{1}{2}+\frac{1}{2} - \frac{1}{3}+\frac{1}{3} - .... = 1$.

8. So if I want to know the sum of a telescoping I have to express a few terms and see how they cancel?

9. The series $\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right)}$ is very nice example of a collapsing sum.
A partial sum of the series looks this:
$\displaystyle S_K = \sum\limits_{n = 1}^K {\left( {\frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right) = } \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{6}} \right) + \left( {\frac{1}{6} - \frac{1}{8}} \right) \cdots + \left( {\frac{1}{{2K}} - \frac{1}{{2K + 2}}} \right)$$\displaystyle = \left( {\frac{1}{2} - \frac{1}{{2K + 2}}} \right)$
Now it simple to see that $\displaystyle S_K \to \frac{1}{2}$,
Therefore the series converges,

10. I quote myself and reask the quesiton
Originally Posted by ascii
the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?
Is it always like that?

11. Originally Posted by ascii
I quote myself and reask the quesiton Is it always like that?
It is always dangerous to answer such a general question!
However, it is always true that if the sequence of partial sums converges then the series converges and the sum of the series is the limit of the sequence of partial sums.

12. Thank you.

13. Hello,

Be very careful !

The sum of a divergent series and a convergent series is indeed divergent.

But the sum of 2 divergent series is not necessarily divergent.
------------------
Another way of doing it :

$\displaystyle \frac 12 \left(\sum_{n=1}^\infty \tfrac 1n-\sum_{n=1}^\infty \tfrac1{n+1}\right)$

We can transform the second one by changing the indice :
$\displaystyle \sum_{n=1}^\infty \tfrac1{n+1}=\sum_{n={\color{red}2}}^\infty \tfrac 1n$

So the sum is now :
$\displaystyle \frac 12 \left(\sum_{n=1}^\infty \tfrac 1n-\sum_{n=2}^\infty \tfrac 1n\right)=\frac 12 \left(1+\underbrace{\sum_{n=2}^\infty \tfrac 1n-\sum_{n=2}^\infty \tfrac 1n}_{=0}\right)=\frac 12$

14. can i also react?

Originally Posted by ascii
This is the series:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}$

So this is what I did:

$\displaystyle \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= \tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}$
So the first term diverges because it is harmonic and thus the hole series diverges.

I have this marked as wrong, but I don't understand why. Can somebody explain it please?
you cannot do this: $\displaystyle \sum (a_n + b_n) = \sum a_n + \sum b_n$ unless $\displaystyle \sum a_n$ and $\displaystyle \sum b_n$ both converge.

15. Originally Posted by kalagota
can i also react?

you cannot do this: $\displaystyle \sum (a_n + b_n) = \sum a_n + \sum b_n$ unless $\displaystyle \sum a_n$ and $\displaystyle \sum b_n$ both converge.
Why ?

$\displaystyle \sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots$

$\displaystyle \sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots$

rearranging the terms won't change the convergence or divergence, so :

$\displaystyle \sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)$

The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?

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