A general formula for the sum of the series 1/k?

In summary, there is no general expression for the sum to the nth term of the series 1/k, but a good approximation for sufficiently large indices is ln(b/a) where b and a are the upper and lower limits. The integral of the function 1/x can also be used to calculate the sum, and it can be expressed as a harmonic number. The harmonic numbers are defined as the sum of 1/k from k=1 to n. The formula for the harmonic number is also shown to be equal to the summation form and the integral form, providing a more satisfactory answer.
  • #1
s0ft
83
0
Is there any general expression for the sum to nth term of the series 1/k?

I know that for sufficiently large indices, a good approximation can be ln(b/a) where b and a are the upper and lower limits respectively.
I've tried to do something very simple for the exact sum by constructing unit length rectangles between the curve of the function 1/x and the x axis. This shows clearly that as the values of x gets larger and larger, the integral, being the portion of the rectangle below the curve, gets closer and closer to the actual sum, the total area of the rectangles.
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But I don't seem to be able to get any further than this. I did try to account for the extra (above-curve) portions of the rectangles by treating them approximately as triangles(what better option did I have?) and using the 0.5*b*h formula for their area but later realizing that the heights are changing, I thought it wouldn't be so easy.
So, anyone willing to give it a try?
 
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  • #3
I couldn't understand wikipedia pwsnafu. Care to explain?
 
  • #4
s0ft said:
I couldn't understand wikipedia pwsnafu. Care to explain?

Sure. The harmonic numbers are defined ##H_n = \sum_{k=1}^n \frac{1}{k}##. Now obviously when you calculate your sum, you'll get a rational number. Fun little fact, the numerators you get (in order) is called the Wolstenholme numbers, and the denominators are called...I don't think they have a name. Hmm.

Anyway, you want to calculate ##\sum_{k=a}^b \frac{1}{k}## when that is ##\sum_{k=1}^b \frac{1}{k} - \sum_{k=1}^{a-1} \frac{1}{k} = H_{b}-H_{a-1}##. So that's the answer, you just look them up on the internet and we are done.

But that's not really a satisfactory answer. Now Wikipedia shows two other forms:
##H_n = \int_0^1 \frac{1-x^n}{1-x} dx = \sum_{k=1}^{n}(-1)^k \frac{1}{k} {n \choose k}##.
The first equality is easy: we know
##\frac{1-x^n}{1-x} = \frac{(1-x)(1+x+x^2+\ldots+x^{n-1})}{1-x} = 1+x+x^2+\ldots+x^{n-1}##.
Integrating term by term we get
##x+\frac{x^2}{2}+\frac{x^3}{3} + \ldots + \frac{x^n}{n}##. The terms of the integral is just x=1 and x=0, so we obtain ##H_n##.

Follow?
 
  • #5
I got lost from the integral for Hn.
Even if I take it as a given, I don't get how it equals the summation part.
 
  • #6
s0ft said:
I got lost from the integral for Hn.
Even if I take it as a given, I don't get how it equals the summation part.

But you do follow what I wrote right?

So moving onto the summation. We want to solve ##\int_0^1 \frac{1-x^n}{1-x}dx##.
We do the substitution ##u = 1-x##. We get
##\int_0^1 \frac{1-(1-u)^n}{u} du##
if this is not immediately obvious verify it.

Now we expand. You know from the binomial theorem that
##(1-u)^n = 1 - u + {n \choose 2} u^2 - \ldots (-1)^n u^n##
So
##1-(1-u)^n = u - {n \choose 2} u^2 + \ldots (-1)^{n-1} u^n##
And
##\frac{1-(1-u)^n}{u} = 1 - {n \choose 2} u + \ldots (-1)^{n-1} u^{n-1}##

So we calculate
##\int_0^1 1 - {n \choose 2} u + \ldots (-1)^{n-1} u^{n-1} \, du##
is equal to ##u - {n \choose 2} \frac{u^2}{2} + \ldots (-1)^{n-1} \frac{u^n}{n} |_0^1##
and we are left with
##1 - {n \choose 2} \frac{1}{2} + \ldots (-1)^{n-1} \frac{1}{n}##
and that's exactly the sum formula.

Edit: 500th post. Yay!
 
  • #7
Thank you very much!
 

1. What is the general formula for the sum of the series 1/k?

The general formula for the sum of the series 1/k is S = 1 + 1/2 + 1/3 + 1/4 + ... + 1/n = ln(n) + γ, where n is the number of terms in the series and γ is the Euler-Mascheroni constant.

2. How do you derive the general formula for the sum of the series 1/k?

The general formula for the sum of the series 1/k can be derived using the Euler-Maclaurin summation formula, which involves approximating the sum with an integral and then applying a correction term. This results in the formula S = ln(n) + γ.

3. What is the significance of the Euler-Mascheroni constant in the formula for the sum of the series 1/k?

The Euler-Mascheroni constant, denoted by γ, is a mathematical constant that appears in many mathematical problems involving the natural logarithm. In the formula for the sum of the series 1/k, it is used to account for the difference between the sum and the integral approximation, resulting in a more accurate formula.

4. Can the general formula for the sum of the series 1/k be applied to infinite series?

Yes, the general formula for the sum of the series 1/k can be applied to infinite series as long as the series converges. However, the value of the series may approach infinity as n increases, so the formula should be used with caution in these cases.

5. Are there any other methods to find the sum of the series 1/k besides the general formula?

Yes, there are other methods to find the sum of the series 1/k, such as using a computer program or calculator to calculate the sum numerically. Additionally, there are some special cases where the sum can be found using other mathematical techniques, such as telescoping series or geometric series.

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