# Sequence equation. Looking for a value at n = infinity. Basic stuff. [duplicate]

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?

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## marked as duplicate by MJD, user127.0.0.1, Yiorgos S. Smyrlis, user1337, ThomasFeb 9 '14 at 18:06

The harmonic series $\;1+\frac12+\frac13+\ldots+\frac1n+\ldots\;$ diverges, meaning it has no finite sum. – DonAntonio Feb 9 '14 at 12:58
"infinity" is not a number. Google "Harmonic series". – David Mitra Feb 9 '14 at 12:58
What is the sum then? Is it possible to calculate? – Unknown Feb 9 '14 at 13:07
What part of "it has no finite sum" you didn't understand, @Unknown ? – DonAntonio Feb 9 '14 at 13:11
oh sorry I thought i read infinite instead of finite. thanks for the help – Unknown Feb 9 '14 at 13:12

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 $\lim_{n \rightarrow \infty} S_n = \infty$.