Proving the Harmonic Series Sum Formula for Positive Integers | Math Proof

[James889]

Hai,

The harmonic series is given by: $$H_{n} = \sum_{i=1}^n \frac{1}{i}$$

I need to prove that for all positive integers:
$$\sum_{j=1}^n H_{j} = (n+1)H_{n} -n$$

So i have
$$H_{5} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{137}{60}$$

$$H_{5} \neq (5+1)*\frac{137}{60} -5$$

Have i missed something here?

Please excuse my epic fail math skills...

 

[Office_Shredder]

So for your H₅ example what they want you to sum is

H₁ + H₂ + H₃ + H₄ + H₅

When I did that I got what the problem tells you you will get

 

[Architeuthis Dux]

Hi,

Don't worry, your math skills are not an epic fail! The proof you have provided is actually correct. The formula you are trying to prove is known as the "Euler-Mascheroni constant" and it can be written as:

\gamma = \lim_{n\to\infty} \left(\sum_{i=1}^n \frac{1}{i} - \ln(n)\right)

In other words, the sum of the harmonic series is equal to the Euler-Mascheroni constant plus the natural logarithm of n. This is a well-known result in mathematics and has been proven by many mathematicians over the years.

One way to prove this formula is by using the technique of mathematical induction. This involves showing that the formula holds for the first few values of n, and then proving that if it holds for a particular value of n, it also holds for the next value of n. This can be done in a step-by-step manner until we reach the general case of n.

In your case, you have shown that the formula holds for n=5, which is the first few values of n. Now, to prove that it holds for the next value of n, we can use the fact that:

H_{n+1} = H_n + \frac{1}{n+1}

Substituting this into our formula, we get:

\sum_{j=1}^{n+1} H_j = (n+1)H_n -n + H_{n+1}

Expanding this further, we get:

\sum_{j=1}^{n+1} H_j = (n+1)H_n -n + H_n + \frac{1}{n+1}

Using the formula for H_n, we get:

\sum_{j=1}^{n+1} H_j = (n+1)H_n -n + (n+1)H_n -n + \frac{1}{n+1}

Simplifying this, we get:

\sum_{j=1}^{n+1} H_j = (n+1)H_{n+1} - (n+1)

Which is the same as the formula we are trying to prove. Therefore, we have shown that if the formula holds for n, it also holds for n+1. By the principle of mathematical induction, this means