Odd - Even?

Calculus Level 3

Suppose that

$A = \sum_{i=0}^\infty \left( \frac{1}{ 2i+1} - \frac{1} { 2i + 2 } \right).$

What is the value of

$\sum_{i=0}^\infty \left(\frac{ 1 } { 4i+1} + \frac{1}{ 4i+3} - \frac{1}{2i+2} \right)?$

\frac{3}{2} A

2A

A

0

2 solutions

Calvin Lin Staff
Nov 22, 2015

For those who wish to complain that "Aren't we just finding the alternating harmonic sum? Hence the answer is A?", it is not necessarily true that we can arbitrarily rearrange a sequence and still get the same final sum. That is only true if the sequence is absolutely convergent, which the harmonic sequence is not.

To solve this problem, you will need the following fact:

If $\sum a_i$ and $\sum b_i$ are convergent, then $\sum (a_i + b_i)$ is convergent and equal to $\sum a_i + \sum b_i$ .

On to the solution:

Consider the sequence $a_i = \frac{ 1}{ 4i+1} - \frac{1}{ 4i+2} + \frac{1}{ 4i+3 } - \frac{1}{ 4i + 4 }$ . We are given that $\sum a_i = A$ .
Consider the sequence $b_i = \frac{1}{ 2 (2i + 1) } - \frac{1}{ 2 ( 2i+2) }$ . We are given that $\sum b_i = \frac{A}{2}$ .

We have $(a_i + b_i) = \frac{1}{ 4i+1} + \frac{1}{ 4i + 3 } - \frac{2}{ 4i+4}$ (after simplification), which is exactly the sequence that we are interested in. Hence, by the above fact, $\sum (a_i + b_i) = \sum a_i + \sum b_i = A + \frac{A}{2} = \frac{3}{2} A$ .

Thanks for this solution. Now I see the merit in your statement.

Rajen Kapur - 5 years, 6 months ago

Sequences that are not Absolutely Convergent can be confusing when we first encounter them. It's goes against (initial) intuition that rearranging the order of terms could lead to a different result, whereas that's not true in the finite case.

This is one of my favorite sequences to play around with. Here's another claim that you should try to prove:

For any real number $r$ , there exists a bijection $a_r ( \cdot): \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ such that

$\sum_{i=1}^{\infty} \frac{ (-1)^{(a_r (i) + 1) }} { a_r (i) } = r.$

This shows that "any number can be written as the limit of the harmonic series when suitably rearranged.

Calvin Lin Staff - 5 years, 6 months ago

What should I use to prove the following $\displaystyle\lim_{n \to \infty} \displaystyle\sum_{i=0}^{n} \frac{1}{2(i+n)+3}=\ln \sqrt{2}$ .

Mahdi Al-kawaz - 5 years, 6 months ago

I believe that the typical Riemann Sums trick would work, since the limiting value $\approx \int_{n}^{2n} \frac{1}{2x+3} \, dx \approx \frac{1}{2} \ln 2$ .

Calvin Lin Staff - 5 years, 6 months ago

Many thanks, spent 5 hours on the internet searching for sum of infinite infinitely small values.

So is this correct ?(still don't -fully- understand the convergent series)

$S_{n}=\displaystyle\sum_{i=0}^{n} ( \frac{1}{4i+3}+\frac{1}{4i+1}-\frac{1}{2i+2} ) = \displaystyle\sum_{i=0}^{n} ( \frac{1}{2i+1} -\frac{1}{2i+2} ) + \displaystyle\sum_{i=0}^{n} \frac{1}{2(i+n)+3}$

Hence

$\displaystyle\lim_{n \to \infty} S_{n} = \ln 2 + \frac{1}{2} \ln 2 = A + \frac{A}{2}$

Mahdi Al-kawaz - 5 years, 6 months ago

@Mahdi Al-kawaz – Yes, that is one possible approach to the question. It requires you to
1. Know that $A = \ln 2$ ,
2. Know how to evaluate $\sum \frac{1}{ 2 ( i+n) + 3 }$ .

You can see my solution, for how to approach this problem without knowing either of these facts. (I did not intend for the problem to be done according to your approach.)

Calvin Lin Staff - 5 years, 6 months ago

@Calvin Lin – The last term in $a_{i}$ isn't it suppose to be negative?

Mahdi Al-kawaz - 5 years, 6 months ago

@Mahdi Al-kawaz – Yes, edited, thanks!

Calvin Lin Staff - 5 years, 6 months ago

Notwithstanding the correctness of the fact stated above, both the sequences are indeed equal to ln 2. No way you can justify the multiplier 3/2 to conclude the answer as 3A/2.

Rajen Kapur - 5 years, 6 months ago

That's not true. With sequences that are not absolutely convergent, and the terms converge to 0, there are ways for us to rearrange the order of terms, to get any arbitrary value that we want.

For example, I can rearrange the alternating harmonic series, to obtain a value greater than 1 in the following way:

Start with +1
To $- \frac{1}{2}$ , we want to add more odd terms so that we get a positive value. In this case, we use the term $- \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
To $- \frac{1}{4}$ , we want to add more odd terms so that we get a positive value. In this case, we use the term $- \frac{1}{4} + \frac{1}{7} + \frac{1}{9}$
To $- \frac{1}{6}$ , we want to add more odd terms so that we get a positive value. In this case, we use the term $- \frac{1}{6} + \frac{1}{11} + \frac{1}{13}$
More generally, to the term $- \frac{1}{2n}$ , we pair it up with the term $+ \frac{1}{4n-1} + \frac{1}{4n + 1 }$ .
In this way, we get 1 + something positive + something positive + something positive + ..., which is clearly greater than 1.

Thus, this explains why

$\begin{aligned} & (\frac{1}{1} + \frac{1}{3} - \frac{1}{2}) + (\frac{1}{5} +\frac{1}{7} - \frac{1}{4} ) + (\frac{1}{9} +\frac{1}{11} - \frac{1}{6} ) + \ldots \\ = & \frac{1}{1} + (\frac{1}{3} - \frac{1}{2} + \frac{1}{5}) +(\frac{1}{7} - \frac{1}{4} + \frac{1}{9}) +(\frac{1}{11} - \frac{1}{6} + \frac{1}{13} + \ldots \\ > & 1 \end{aligned}$

Now, it remains to guess if the value is $\frac{3}{2} \ln 2 \approx 1.04$ , or $2 \ln 2 \approx 1.39$ .

Calvin Lin Staff - 5 years, 6 months ago

Both the sequences in the question are same as $\sum_{n=1}^{\infty} \left( -1 \right)^{n - 1} \frac{1}{n}= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +\ldots =ln 2$ indubitable summation of a convergent series. I do not think by repackaging the sequence differently a convergent series like above converges to different values.

Rajen Kapur - 5 years, 6 months ago

@Rajen Kapur – No, they are not the same. For a non-absolutely convergent sequence, the order of the terms matters, especially when terms are moved "very far away". As shown above, repackaging the sequence does result in a different value.

Calvin Lin Staff - 5 years, 6 months ago

@Rajen Kapur – People who don't see shall get it right but people who see can get it wrong. You try to evaluate the exact sum using Excel. Then you shall find that what is said is true. Just spend few minutes to try.

Lu Chee Ket - 5 years, 6 months ago

Mahdi Al-kawaz
Dec 16, 2015

In a less-smart way
$\displaystyle\sum_{i=0}^{\infty} \frac{1}{4i+1}+\frac{1}{4i+3}-\frac{1}{2i+2}\\ =\lim_{n \to \infty} ( \sum_{i=0}^{n} \frac{1}{2i+1}-\frac{1}{2i+2})+\sum_{i=0}^{n} \frac{1}{2(i+n)+3} \\ =A+\frac{1}{2} \lim_{n \to \infty} \sum_{i=0}^{n} \frac{1}{i+n} \\ =A+\frac{1}{2}\lim_{n \to \infty} ( \sum_{i=0}^{2n} \frac{1}{i+1} - \sum_{i=0}^{n} \frac{1}{i+1} )\\ =A+A/2$

Odd - Even?

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