Find Sequence Property & Sum Property of ΣnK=1K^-1

[Wanted]

Is it possible to come up for a sequence property or sequence sum property for Σⁿ_K=1K^-1

If so, what other sequence properties that are not commonly seen are there?

 

[BvU]

Well, it is known that $$\ln (1-x) = - \Sigma_{n=1}^\infty \ {x^n\over n} $$ for $|x|<1$ , so you have a diverging series here. I think I remember the progress of the sum is logarithmic, but I can't reproduce that offhand. Had to do with changing over from sum to integral for $n$ large enough...

 

[Samy_A]

Not sure I understand the question, but might this help?

 

[Mark44]

Is it possible to come up for a sequence property or sequence sum property for Σⁿ_K=1 K^-1

What do you mean by "sequence property" and "sequence sum property"? These are not terms that are commonly used with sequences or series.

 

[Wanted]

Not sure I understand the question, but might this help?

The link helps yes

The question is simple... is there a formula (or possible to create one) for calculating k^-1 like there is for K^1, k^2 etc.

Editedow to find the sum and not just the nth term hmm

What do you mean by "sequence property" and "sequence sum property"? These are not terms that are commonly used with sequences or series.

Not commonly used?

Some basic commonly used terms in pre-calculus

 

[BvU]

Finding the nth Term it looks like

Nope, that is from 1 to n !

 

[Mark44]

The link helps yes

The question is simple... is there a formula (or possible to create one) for calculating k^-1 like there is for K^1, k^2 etc.

You're not "calculating k^-1". The summation on the right side is the closed form of the sum in the middle.

Finding the nth Term it looks like

or just ln(n)

The integral above is an approximation of the sum of the finite series H_n. The integral does not give you the n^th term. The n^th term of the sum is just 1/n.

but how to find the sum and not just the nth term hmmNot commonly used?

I thought you meant specific properties of the finite sum you showed in post #1.

Some basic commonly used terms in pre-calculus

The above are really properties of finite sums, not sequences (despite the title).

 

[Wanted]

You're not "calculating k^-1". The summation on the right side is the closed form of the sum in the middle.
The integral above is an approximation of the sum of the finite series H_n.

I know this.. I'm using the image to [better] illustrate again the equation in question

The integral does not give you the n^th term. The n^th term of the sum is just 1/n.

True

I thought you meant specific properties of the finite sum you showed in post #1.

The above are really properties of finite sums, not sequences (despite the title).

Semantics aside it should be obvious considering the original post shows the Sigma notation. The actual question here still persists

 

[Mark44]

Semantics aside it should be obvious considering the original post shows the Sigma notation. The actual question here still persists

Your question (quoted below) included a summation, but asked about properties of a sequence, which suggests that you aren't clear about the differences between a sequence and a sum. Your question was very unclear, so you can't just brush things off with "semantics aside".

Here is what you wrote in post 1.

Is it possible to come up for a sequence property or sequence sum property for Σⁿ_K=1K^-1

If so, what other sequence properties that are not commonly seen are there?

If you have a sum such as $a_1 + a_2 + \dots + a_n$, you can always write this in closed form as $\sum_{i = 1}^n a_i$, but it might not be possible to write the associated infinite series as a function.

 

[Wanted]

Your question (quoted below) included a summation, but asked about properties of a sequence, which suggests that you aren't clear about the differences between a sequence and a sum. Your question was very unclear, so you can't just brush things off with "semantics aside".

Despite them having similar connotations and being shown explicitly it somehow remains vague?

it might not be possible to write the associated infinite series as a function.

n is denoted above the sigma and not infinity implying that it is a finite series... considering the alternative "might not be possible" makes for a poor interpretation of what was meant.

Could we please veer off the subject of semantics and focus on the arithmetic, we both clearly understand each-other, at least now. None of us are on some math standardizing committee that will get to decide any of this anyways nor is it of my own interest to discuss

 

[Mark44]

Your question (quoted below) included a summation, but asked about properties of a sequence, which suggests that you aren't clear about the differences between a sequence and a sum. Your question was very unclear, so you can't just brush things off with "semantics aside".

Despite them having similar connotations and being shown explicitly it somehow remains vague?

A sequence and a sum (finite or infite) are two different things. I don't know what you mean by "similar connotations."
Here is a sequence: $\{1, 1/2, 1/4, 1/8, ...\}$
Here is a (finite) series: $\sum_{i = 0}^n \frac 1 {2^n}$
The terms in a series come from a sequence, but the two concepts are very different.

it might not be possible to write the associated infinite series as a function.

For certain kinds of series, such as geometric series (as in the example above), there are formulas for writing the sum in closed form. For an arbitrary series, there is no method that I'm aware of to allow rewriting the series in closed form.

n is denoted above the sigma and not infinity implying that it is a finite series... considering the alternative "might not be possible" makes for a poor interpretation of what was meant

Could we please veer off the subject of semantics and focus on the arithmetic, we both clearly understand each-other, at least now. None of us are on some math standardizing committee that will get to decide any of this anyways nor is it of my own interest to discuss

A question has meaning; i.e., semantic content. If the meaning of the question is not clear, it's difficult for anyone to answer the question.

 

[pwsnafu]

Euler showed
$\sum_{k=1}^n \frac{1}{k} = \int_{0}^{1} \frac{1-x^n}{1-x} \, dx = \sum_{k=1}^n (-1)^k \frac{1}{k} {n \choose k}$
also we have
$H_n = \psi(n) + \gamma$
where $\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.

 

[Wanted]

Euler showed
$\sum_{k=1}^n \frac{1}{k} = \int_{0}^{1} \frac{1-x^n}{1-x} \, dx = \sum_{k=1}^n (-1)^k \frac{1}{k} {n \choose k}$
also we have
$H_n = \psi(n) + \gamma$
where $\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.

What is the derivative of

Is it just ln(T(n)) ?

so

$\sum_{k=1}^n \frac{1}{k} = ln((n-1)!)/((n-1)!) + \gamma$ ??

Yea plugging in the numbers doesn't work. Is that the correct derivative?

 

[Samy_A]

What is the derivative of

Is it just ln(T(n)) ?

so

$\sum_{k=1}^n \frac{1}{k} = ln((n-1)!)/((n-1)!) + \gamma$ ??

Yea plugging in the numbers doesn't work. Is that the correct derivative?

No.

I'm not sure what you mean by T(n), but as far as I know the best we have is what @pwsnafu already posted:

$H_n = \psi(n) + \gamma$
where $\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.

 

[Svein]

The standard definition of the Euler-Mascheroni constant is: \gamma = \lim_{n\rightarrow \infty}( (\sum_{k=1}^{n}\frac{1}{k})-\ln(n)). Rearrange this, and you have \sum_{k=1}^{n}\frac{1}{k}\approx\ln(n)+\gamma.

This expression can be manipulated in various ways, I shall just point out one result: In order for \sum_{k=1}^{n}\frac{1}{k}&gt;100, you need \ln(n)\geq 100 approximately. This means that n\geq e^{100}\approx 2.7\cdot 10^{43}. So, while the sum diverges, it diverges very slowly.

 

[Wanted]

No.

I'm not sure what you mean by T(n), but as far as I know the best we have is what @pwsnafu already posted:

T(n) https://en.wikipedia.org/wiki/Gamma_function
T'(n) https://en.wikipedia.org/wiki/Digamma_function https://en.wikipedia.org/wiki/Polygamma_function

I believe I interpreted the first derivative T' wrong though in the above post

The standard definition of the Euler-Mascheroni constant is: \gamma = \lim_{n\rightarrow \infty}( (\sum_{k=1}^{n}\frac{1}{k})-\ln(n)). Rearrange this, and you have \sum_{k=1}^{n}\frac{1}{k}\approx\ln(n)+\gamma.

This expression can be manipulated in various ways, I shall just point out one result: In order for \sum_{k=1}^{n}\frac{1}{k}&gt;100, you need \ln(n)\geq 100 approximately. This means that n\geq e^{100}\approx 2.7\cdot 10^{43}. So, while the sum diverges, it diverges very slowly.

I'm less interested in the infinite (divergent) series. Although I know in some parts of science they tend to try and set values for infinite (divergent) series if I remember correctly the sum of all natural numbers is understood as -1/16... but this isn't what interests me.

I'm more interested in the finite series because I don't want to have to write out or use calculators to find the sum of something like
\sum_{k=1}^{n}\frac{1}{k}.

I tried plugging in the value for something like

\sum_{k=1}^{4}\frac{1}{k}. and the formula

which is exactly 2.083333...33 or 25/12 (using expansion and exhaustion)

\sum_{k=1}^{4}\frac{1}{k}\approx\ln(4)+\gamma.

is about 1.9635

So although this looks promising and useful (to an extent) it is not exact (within significant variance) and not exactly what I'm looking for but very interesting and a step in the right direction for sure.

 

[Samy_A]

T(n) https://en.wikipedia.org/wiki/Gamma_function
T'(n) https://en.wikipedia.org/wiki/Digamma_function https://en.wikipedia.org/wiki/Polygamma_function

I may be missing something here, but I still don't see what T(n) should be.
If you mean $\Gamma (n)$, that is a little confusing.

I'm less interested in the infinite (divergent) series. Although I know in some parts of science they tend to try and set values for infinite (divergent) series if I remember correctly the sum of all natural numbers is understood as -1/16... but this isn't what interests me.

I'm more interested in the finite series because I don't want to have to write out or use calculators to find the sum of something like
\sum_{k=1}^{n}\frac{1}{k}

Finding such a formula would be a notable achievement, not sure it is possible.

 

[Wanted]

I may be missing something here, but I still don't see what T(n) should be.
If you mean $\Gamma (n)$, that is a little confusing.

It's confusing to me as well, but it's used in the above function he posted:

$H_n = \psi(n) + \gamma$
where $\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}$ is the digamma function

T and T' are used which the article about the digamma function refers to as this

The gamma function

But still confused on what that first derivative of it is $\Gamma '(n)$ .. since I don't actually know how to take derivatives of extremes.

Finding such a formula would be a notable achievement, not sure it is possible.

I think it would be too, both notable and potentially impossible.. at least with standard notation or even any notation. Unless someone has done it already, which would be nice to learn about as well.

 

[Samy_A]

It's confusing to me as well, but it's used in the above function he posted:

$H_n = \psi(n) + \gamma$
where $\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}$ is the digamma function

T and T' are used which the article about the digamma function refers to as this

I really don't want to start about semantics again, but I still don't understand. Nowhere in the linked Wikipedia pages do I see a function T being used. Could you simply define this function to make things clear (for me, maybe I'm missing the obvious).

But still confused on what that first derivative of it is $\Gamma '(n)$ .. since I don't actually know how to take derivatives of extremes.

The derivative of the $\Gamma$ function has been given by @pwsnafu in terms of the digamma function. No idea what "take the derivative of extremes" means.
There is (as far as I know) no formula for $\Gamma '(n)$ using only "elementary" elements. See this thread for a integral formula for $\Gamma'(z)$.

 

[Wanted]

I really don't want to start about semantics again, but I still don't understand. Nowhere in the linked Wikipedia pages do I see a function T being used. Could you simply define this function to make things clear (for me, maybe I'm missing the obvious).

For the love of all that is Euclidean I sincerely hope you're not talking about T(n) vs $\Gamma (n)$ considering one was just used over the other to avoid having to learn how to/use the forum API etc.

It is at the very top of this page https://en.wikipedia.org/wiki/Gamma_function

The derivative of the $\Gamma$ function has been given by @pwsnafu in terms of the digamma function. No idea what "take the derivative of extremes" means.
There is (as far as I know) no formula for $\Gamma '(n)$ using only "elementary" elements. See this thread for a integral formula for $\Gamma'(z)$.

Derivative of extemems i.e. f(x)=(x-1)! what is f'(x)

The conclusion in that thread "cannot be written in simpler ways"
That's unfortunate, if it's entirely true -- at least for this application. Since that would mean it's basically useless at this point to use to find a summation formula for the finite series of

\sum_{k=1}^{n}\frac{1}{k}.

 

[Samy_A]

For the love of all that is Euclidean I sincerely hope you're not talking about T(n) vs $\Gamma (n)$ considering one was just used over the other to avoid having to learn how to/use the forum API etc.

Mathematics starts with clear notation and precise definitions. I indeed was talking about T(n) vs $\Gamma (n)$, but that has been cleared up now, thanks.

The conclusion in that thread "cannot be written in simpler ways"
That's unfortunate, if it's entirely true -- at least for this application. Since that would mean it's basically useless at this point to use to find a summation formula for the finite series of

\sum_{k=1}^{n}\frac{1}{k}.

That's probably true.

 

[Svein]

I'm more interested in the finite series because I don't want to have to write out or use calculators to find the sum of something like
∑nk=11k\sum_{k=1}^{n}\frac{1}{k}.

I tried plugging in the value for something like

∑4k=11k\sum_{k=1}^{4}\frac{1}{k}. and the formula

which is exactly 2.083333...33 or 25/12 (using expansion and exhaustion)

∑4k=11k≈ln(4)+γ\sum_{k=1}^{4}\frac{1}{k}\approx\ln(4)+\gamma .

is about 1.9635

So although this looks promising and useful (to an extent) it is not exact (within significant variance) and not exactly what I'm looking for but very interesting and a step in the right direction for sure.

1. The approximation is only good for large values of n (at least in the thousands).
2. There is no known closed formula for the sum
3. An old treatise on numeric mathematics (updated by me) suggests: for n<100, just do the sum (use Excel). For n≥100, a good approximation is s_{n}=s_{100}+\int_{100}^{n}\frac{1}{x}dx=s_{100}+\ln(n)-\ln(100).

 

[Wanted]

1. The approximation is only good for large values of n (at least in the thousands).
2. There is no known closed formula for the sum
3. An old treatise on numeric mathematics (updated by me) suggests: for n<100, just do the sum (use Excel). For n≥100, a good approximation is s_{n}=s_{100}+\int_{100}^{n}\frac{1}{x}dx=s_{100}+\ln(n)-\ln(100).

1. Yea I noticed that it gets more accurate the larger the value is, so it could be quite useful depending on the application/size of the series.
2. Hopefully we can find one
3. Well yea but the whole point was to find a formula, using excel to find the sum would kind of defeat the point :p

 

[zinq]

I don't think I've ever seen a closed form formula for the nth harmonic number:

H_n = 1 + 1/2 + 1/3 + ... + 1/n.
​

But many things are known about it. As has been mentioned, its difference from ln(n) approaches the finite limit known as γ.

It's also known that for n > 1, H_n is never equal to an integer.

A formula for H_n (that may not be what you're hoping for) is as follows:
$$H_n = \int\limits_{0}^{1} \frac{1-x^n}{1-x}dx$$

An interesting aspect of this equation is that it makes sense even when n is not an integer!

So in fact there is a natural definition of H_s for any s ≥ 0.