Sequence equation. Looking for a value at n = infinity. Basic stuff.

Question

I don't know how to do sigma equations, but my question is:

1/1 + 1/2 + 1/3 + 1/4 + 1/5.. .. + 1/infinity = ?

can this be calculated? is the answer infinity or does it stop at a value?

I tried to create the sequence in c# (very simple program).

        double x = 0;
        double n = 1;
        while (true)
        {
            x = x + 1 / n;
            Console.WriteLine(x);
            i++;
        }

cmd is capable of calculating 10k sequences per second. I can see that the value is slowing down. At stage 10.000.000 the value is about 16,7.

How much is the value at n=infinity?

The harmonic series $;1+\frac12+\frac13+\ldots+\frac1n+\ldots;$ diverges, meaning it has no finite sum. — DonAntonio, Feb 09 '14 at 12:58
What part of "it has no finite sum" you didn't understand, @Unknown ? — DonAntonio, Feb 09 '14 at 13:11
oh sorry I thought i read infinite instead of finite. thanks for the help — Unknown, Feb 09 '14 at 13:12
If you stop after $n$ summands, the result is $\approx \ln n +\gamma-\frac1{2n}$ where $\gamma\approx 0.5772$. Note that $\ln n\to \infty$ for $n\to \infty$. — Hagen von Eitzen, Feb 09 '14 at 13:25

score 2 · Answer 1 · answered Feb 09 '14 at 13:08

Let's call $S_n = \sum_{i=1}^{n} \frac{1}{i}$. $S_{2n} = \sum_{i=1}^{2n} \frac{1}{i}$.

If $S_n$ converges, then $S_{2n}$ would converge to the same limit.

However, $S_{2n}-S_{n} = \sum_{i=n+1}^{2n} \frac{1}{i} \geq \sum_{i=n+1}^{2n} \frac{1}{2n} = n*\frac{1}{2n} = \frac{1}{2}$

As a consequence, $S_{n}$ does not converge and $\lim_{n \rightarrow \infty} S_n = \infty$.

Sequence equation. Looking for a value at n = infinity. Basic stuff.

1 Answers1