As i see it, the Harmonic series is a limit of a sum. It is not just \(\displaystyle \lim_{n\to {\infty}}\frac{1}{n}\).
It is the sum of the first n reciprocals as n-->infinity. See what I am getting at?.
At least, I think that is what Hank means. Maybe I am wrong.
\(\displaystyle =\frac{-1}{2}\sum_{k=1}^{n}\frac{1}{k}=\frac{-1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{n}\right)\)
\(\displaystyle -\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+....................\frac{1}{n}\right)\)
But, as n gets larger and larger, we get the gamma function:
\(\displaystyle \frac{-1}{2}\lim_{n\to {\infty}}\sum_{k=1}^{n}=\frac{-1}{2}\left({\gamma}+ln(n+1)\right)\)
Try a value of n and see. As it gets larger and larger, it approaches said result.
\(\displaystyle \frac{-1}{2}\sum_{k=1}^{100}=-2.5936887588......\)
Now, plug into \(\displaystyle \frac{-1}{2}\left(.577+ln(101)\right)=-2.5960602584...\)
Just a fun result from the gamma function.
As n gets larger and larger, we head into negative territory and the limit of the sum is \(\displaystyle -{\infty}\)
Which stands to reason, since we know the Harmonic series diverges. This one has a negative tacked on the front , so it 'approaches' negative infinity.
I do not like to say anything 'approaches' infinity. That is why I wrapped it in quotes.
Also, have you heard of the Psi function?. Technically, we get \(\displaystyle \frac{-1}{2}\lim_{n\to {\infty}}\sum_{k=1}^{n}\frac{1}{k}=\frac{-1}{2}\lim_{n\to {\infty}}({\Psi}(n+1)+{\gamma})={-\infty}\)
