How would you solve this?

[limpid pond]

|| $\lim_{x \to \infty}\left( \displaystyle\sum_{k = 1}^{x}\dfrac{1}{k} - \ln{x} \right) = \gamma$. ||

[solemn kraken]

Killuminati

[limpid pond]

Where || $\overline{\mathbb{Q}} \backepsilon \gamma \approx 0.58$ is the Euler-Mascharoni constant. ||

[solemn kraken]

Killuminati

[limpid pond]

Use this as a starting point. The rest is pretty standard if you've worked with limits.

[astral wraith]

Hi, where do you get these problems from??

[patent bridge]

Daily Integral

[limpid pond]

There's a website given

[limpid pond]

Do you actually need the rest of the solution?

[astral wraith]

Oh I see.

Thanks.

[limpid pond]

Or do you just post these for fun? @patent bridge ?

[patent bridge]

If the question is interesting

[astral wraith]

Can you put it in spoiler I'd like to try it too, thanks in advance.

[astral wraith]

Yeah no I couldn't do it. 🥀

Ima go to sleep now.

[limpid pond]

I more or less gave you the hardest part of the problem lol. All you had to do was re-write the limit and then identify it with the exponential limit.

It's literally just || e^γ ||.

[astral wraith]

Oh no I was starting from the start lol.

[astral wraith]

I don't get how you got here though.