its convergent will converge to 0 how can it diverge?
as 1-1=0 1/2-1/2=0 and so on
apply ratio test u will get <0 so convergent
refer to summation of 1/n-1/n from 1 to infinity - Wolfram|Alpha for more
its convergent will converge to 0 how can it diverge?
as 1-1=0 1/2-1/2=0 and so on
apply ratio test u will get <0 so convergent
refer to summation of 1/n-1/n from 1 to infinity - Wolfram|Alpha for more
If we define ${a_n} = {\left( { - 1} \right)^{n + 1}}{\left\lfloor {\dfrac{{n + 1}}{2}} \right\rfloor ^{ - 1}}$ then is $\sum\limits_{k = 1}^\infty {{a_k}} $ the same as OP?
If it is look at the sequence of partial sums: ${S_N} = \sum\limits_{k = 1}^N {{a_k}}$.
Now ${S_N} = \left\{ {\begin{array}{*{20}{c}}
{0,}&{\text{n is even}} \\
{{{\left\lfloor {\frac{{n + 1}}{2}} \right\rfloor }^{ - 1}},}&{\text{n is odd}}
\end{array}} \right.$
$\displaystyle{\lim _{N \to \infty }}{S_N} = ?$
From Kaplan: In testing a series of variable signs for convergence, parentheses should not be inserted. Which is what Wolfram is probably using to come up with convergence. This is consistent with the point that I believe Plato is trying to make in defining series as a limit of sequences.
As for Plato's post #5, it is convergent because it is of alternating sign and monotonically decreasing. I don't know what he means by is the sum the same as OP, or it's relevance to OP.
You can add parentheses (grouping) and still have a sequence of partial sums, which include the grouping, but it's just a matter of precise definition.
Given two sequences, one with grouping and one without. Are their sums the same is a legitimate (meaningful) question.
The question of grouping and rearrangement is a little confusing. Can grouping, (parentheses) by considered a rearrangement, since you can get same effect by rearrangement and convention mentioned by Kaplan.
Anyhow, the OP has been answered by the above-mentioned convention. The subject of grouping and rearrangement can lead to a long thread.
Here is the OP.
Note that there are no parentheses in that post. You brought them into the mix.
I gave the series a closed form definition. Now the sequence being summed is alternating but it is not decreasing, it is non-increasing. The sequence of partial sums converges to zero. Therefore, the series sums to zero.
So we have shown the OP series converges to 0.
Then we use the theorem that adding parentheses to a convergent series gives a new convergent series with the same sum. So the question of adding parentheses to OP is academic, but leads to an interesting result:
Consider:
1+(-1+1/2)+(-1/2+1/3)+(-1/3+1/4)+…= 1 - 1/2 - 1/6…. =0 by above theorem.
=1+a1+a2+a3…..=0 where an=(-1/n + 1/(n+1)) = -1/(n(n+1))
It follows that Σ_{1}^{∞}1/(n(n+1))=1
Ref:
adding and removing parentheses in series | planetmath.org
Strange. I received an email notice of a post questioning above link. But it isn't here.
I didn’t go through the article. It was a quick reference to “regrouping doesn’t change sum of a convergent series,” which is easily proved as follows:
Given the convergent series Σan.
Create a new series Σbn by an arbitrary grouping (parentheses) of terms of Σan.
Then every partial sum of Σbn is a partial sum of Σan.
Since the sequence of partial sums of Σan converges, any sub-sequence converges to the same sum.
Referring to my previous post, I note that Σ1/n(n+1) is indeed 1.
That is not correct, The series not be conditionally convergent.
See Riemann's remarkable find here.