Can anyone explain to me this infinite series problem?

[paulfr]

Homework Statement

and in this case we have,

[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/eq0016MP.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/empty.gif

Homework Equations

I can not see how they get either of the 3 terms or where the 3/4 term comes from.
Once given the expression, its easy to see it converges to be 3/4.

The Attempt at a Solution

Even if I limit shift to either i=0 or i=1, I can not see this solution. Wolfram Alpha confirms 3/4 is the correct answer. Thanks for your help.

 

[SammyS]

Homework Statement

and in this case we have,

[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/eq0016MP.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries_files/empty.gif

Homework Equations

I can not see how they get either of the 3 terms or where the 3/4 term comes from.
Once given the expression, its easy to see it converges to be 3/4.

The Attempt at a Solution

Even if I limit shift to either i=0 or i=1, I can not see this solution. Wolfram Alpha confirms 3/4 is the correct answer. Thanks for your help.

Is this from a course?

If so, what is the main topic of coverage?

Try partial fraction decomposition of $\displaystyle \ \frac{1}{i^2-1} \$

 

[paulfr]

Infinite Series convergence and divergence.

PFD will give A / i +1 + B / i - 1 and that does not look in any way like the given solution.

 

[PeroK]

Infinite Series convergence and divergence.

PFD will give A / i +1 + B / i - 1 and that does not look in any way like the given solution.

Work out $A$ and $B$ and keep going!

 

[Mark44]

Infinite Series convergence and divergence.

PFD will give A / i +1 + B / i - 1 and that does not look in any way like the given solution.

Surely you don't mean $\frac A i + 1 + \frac B i - 1$, which is what you wrote.

If you meant $\frac A {i + 1} + \frac B {i - 1}$, you need to at least write this as A/(i + 1) + B/(i - 1); i.e., using paretheses.

 

[vela]

PFD will give A / i +1 + B / i - 1 and that does not look in any way like the given solution.

You shouldn't expect it to because you're comparing apples to oranges. The decomposition is of just 1/(i²-1), but the expression you're trying to derive is for s_n. Note that one expression involves the variable $i$ while the other involves $n$. You have to do a bit more work to connect the two.

 

[paulfr]

Yes, the stated problem says to find the limit for s_n. Can you give me a clue what to do to get on track here ?

 

[vela]

Follow PeroK's advice and keep going.

 

[paulfr]

I did and got A = - 1/2 and B = 1/2 ==> [ - (1/2) / ( i + 1 ) ] + [ (1/2) / ( i - 1 ) ] which does not seem to help. Where is that 3/4 coming from ?
I do not understand what vela is trying to tell me.
I know I am way off base here.

 

[ehild]

I did and got A = - 1/2 and B = 1/2 ==> [ - (1/2) / ( i + 1 ) ] + [ (1/2) / ( i - 1 ) ] which does not seem to help. Where is that 3/4 coming from ?

Write out the sum Sn for n=1, n=2, n=3, n=4... What do you observe?

 

[paulfr]

OK how about an easier question.
Why is limit n=> inf ∑ 1 / n² = π² / 6 ??

When I find the Antiderivative I get - 1/x
Where does the π come from ??

 

[Mark44]

OK how about an easier question.
Why is limit n=> inf ∑ 1 / n² = π² / 6 ??

This isn't actually an easier problem. Using the advice given in this thread are you able to finish the original problem in this thread?

 

[paulfr]

No I can not.
But a p-series is rather basic so if I knew how to do this one
I will likely induce how to do others. First problem is the hardest one.
I know I need to work an improper integral, but can not get to π² / 6

 

[Mark44]

No I can not.
But a p-series is rather basic so if I knew how to do this one
I will likely induce how to do others. First problem is the hardest one.
I know I need to work an improper integral, but can not get to π² / 6

I don't think an improper integral is any help in this problem.

The p-series test tells you whether a series of the form $\sum_{n = 1}^\infty \frac 1 {n^p}$ converges. It doesn't tell you what such a series converges to.

Have you made any progress on the series you first asked about? That one is an example where you can actually find the value the series converges to.

The advice already given in this thread is enough to solve the problem you started with. By decomposing $\frac 1 {i^2 - 1}$ I was able to get exactly what it asks you to show.

 

[SammyS]

I did and got A = - 1/2 and B = 1/2 ==> [ - (1/2) / ( i + 1 ) ] + [ (1/2) / ( i - 1 ) ] which does not seem to help. Where is that 3/4 coming from ?
I do not understand what vela is trying to tell me.
I know I am way off base here.

Write out the sum Sn for n=1, n=2, n=3, n=4... What do you observe?

Similar to @ehild 's advice

Consider n=5:

Write out the four terms obtained from $\displaystyle \ \frac12\frac1{(i-1)}\$ for $\ i =\,$ 2, 3, 4, 5 . Don't bother to sum these.​

Similarly,

Write out the four terms obtained from $\displaystyle \ \frac{-1}2\frac1{(i+1)}\$ for $\ i =\,$ 2, 3, 4, 5 . Don't bother to sum these either.​

Now consider that you need to sum these 8 terms. What pops out at you?.

 

[paulfr]

Sammy S
Thank you.
I can see the 3/4 when I find the 8 terms for n= 2,3,4,5.
It telescopes after 2 iterations of n.
And if the second term of the solution in the OP were not - 1/2n but was
+ 1/ 2(n+1) I would have it understood completely. [Maybe the teacher made an error ?]

I did the Integral Test and got 1/3 not 3/4.
Can not explain that but now I see that the convergence, if there is one, will
be the finite value of the integral.
Still very shaky understanding this but that's what keeps me coming back for more.
Smile
Thanks again.

 

[Mark44]

Sammy S
Thank you.
I can see the 3/4 when I find the 8 terms for n= 2,3,4,5.
It telescopes after 2 iterations of n.
And if the second term of the solution in the OP were not - 1/2n but was
+ 1/ 2(n+1) I would have it understood completely. [Maybe the teacher made an error ?]

No. As written, it is correct. When you expand the summation, write terms for, say, n = 2, 3, 4, 5, as well as terms for n-3, n-2, n-1, and n. You should see that you will have the 3/4 term, but will also have a couple more.

I did the Integral Test and got 1/3 not 3/4.

The integral test just tells you whether the series converges or diverges -- it doesn't give you the sum of the series.

Can not explain that but now I see that the convergence, if there is one, will
be the finite value of the integral.

No. The integral is only loosely related to the summation.

 

[paulfr]

Thanks Mark44
Yes if the integral is finite, the series is convergent.
I did the integral again and got 0.549 as a lower bound
Adding in f(N) = f(2) = 1/3, the upper bound is 0.882 and 3/4 fall between these bounds.
This lower/upper bound information is from Wiki ; Integral Series Test.
Thanks again.

 

[SammyS]

Sammy S
Thank you.
I can see the 3/4 when I find the 8 terms for n= 2,3,4,5.
It telescopes after 2 iterations of n.
And if the second term of the solution in the OP were not - 1/2n but was
+ 1/ 2(n+1) I would have it understood completely. [Maybe the teacher made an error ?]

I did the Integral Test and got 1/3 not 3/4.
Can not explain that but now I see that the convergence, if there is one, will
be the finite value of the integral.
Still very shaky understanding this but that's what keeps me coming back for more.
Smile
Thanks again.

Clearly, you are missing the point.

There is no integral involved.

Had you done what I suggested you would have gotten the following.

Consider n=5:
Write out the four terms obtained from $\displaystyle \ \frac12\frac1{(i-1)}\,,\$ for $i=$ 2, 3, 4, 5 .

$\displaystyle \frac12, \frac14, \frac16, \frac18$​

Write out the four terms obtained from $\displaystyle \ -\frac12\frac1{(i+1)}\,,\$ for $i=$ 2, 3, 4, 5 .

$\displaystyle -\frac16, -\frac18, -\frac1{10}, -\frac1{12}$​

Some terms cancel when you add these. Which ones don't cancel?
.
What happens if n = 12 instead of 5 ?

 

[ehild]

Almost the same as @sammy-s hint...
The sum Sn can be written as $S_n=\sum_2^n={\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)}=\sum_2^n{a_i}$
Write up the terms $a_i=\frac{1}{2}\left(\frac{1}{i-1}-\frac{1}{i+1}\right)$ for i=2, 3, 4, 5...
$a_2=\frac{1}{2}\left(\frac{1}{2-1}-\frac{1}{2+1}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)$
$a_3=\frac{1}{2}\left(\frac{1}{3-1}-\frac{1}{3+1}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)$
$a_4=\frac{1}{2}\left(\frac{1}{4-1}-\frac{1}{4+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)$
$a_5=\frac{1}{2}\left(\frac{1}{4-1}-\frac{1}{4+1}\right)=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)$
What is S₅, for example? You have to sum the a_i-s.
$S_5=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3} + \frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}\right)$
Some terms cancel... Which are the remaining terms? Which are the remaining terms in S_n?