Does this mean that Euler's constant diverges?

Question

I was playing around with the Euler-Mascheroni constant and the constant $\gamma$ is defined as the limit below. $$ \gamma = \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log{n} $$ It can also be expressed as this other sum which is basically the same thing but rewritten. It sums up the individual differences rather than the entire difference. $$ \gamma = \lim_{n\to\infty} \sum_{k=1}^n \left( \frac{1}{k} - \log\left(\frac{k+1}{k} \right) \right) $$ This was where I was playing around and my work is right below. $$ \gamma = \lim_{n\to\infty} \sum_{k=1}^n \left( \frac{1}{k} - \log\left(\frac{k+1}{k} \right) \right) = \lim_{n\to\infty} \sum_{k=1}^n \left( \frac{1}{k} - \log\left(1+\frac{1}{k} \right) \right) \\ =\lim_{n\to\infty} \left[ \sum_{k=1}^n \left( \frac{1}{k} \right) -\sum_{k=1}^n \left( \log\left( 1+\frac{1}{k} \right) \right) \right] $$ And using the properties of logarithms (specifically $\log{ab}=\log{a}+\log{b}$) we can rewrite the left sum as a product. $$ \lim_{n\to\infty} \left[ \sum_{k=1}^n \left( \frac{1}{k} \right) - \log\left( \prod_{k=1}^n \left(1+\frac{1}{k} \right) \right) \right] $$ Actually, while I'm writing this I don't even know why I rewrote the fraction that way because it does nothing but it turns out that that product is equal to $n+1$ (this is easily provable using factorials). So it follows that the limit is equal to $$ \lim_{n\to\infty} \left[ \sum_{k=1}^n \left( \frac{1}{k} \right) - \log(n+1) \right] $$ And in the beginning, $\gamma$ was defined as $$ \gamma = \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log{n} $$ So $$ \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log{n} = \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log(n+1) $$ This can be tested by plugging in some numbers on a calculator like desmos. And this is where things start to break, assuming that my proof is correct. Since this formula holds true for any $n$ at infinity, surely this must be true for $n+2$. And for $n+3$, and $n+l$, where $l$ is a positive integer. And the same thing backward must be true, that is, it's also true for $n-1$ and $n-l$. So what if $l=n-1$? Then the formula yields $$ \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log({n-l}) = \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log({n-(n-1)}) = \lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log({1}) \\ = \sum_{k=1}^n \frac{1}{k} = \infty $$ Does this mean that $\gamma$ isn't even a constant and goes to infinity? I don't think so but I can't really find a mistake in my logic. I don't even know enough calculus to do a test (but I've heard of it) to check if the series converges.

Answer 1

This

$$\lim_{n\to\infty} \sum_{k=1}^n \left(\frac{1}{k}\right) - \log({n-k}) $$doesn't work, at least not the way I think you think it does.

In the definition of $\gamma$, the $\log()$ is outside of the sum, so your $k$ in "$\log(n-k)$" is a free variable in this whole expression. I.e. that's not a constant, it's an expression that depends on $k$.

Further, even after your edit, while it's true you can use "...$-\log(n-m)$" in place of "...$-\log(n)$" for any fixed value of $m$, and the limit will be the same as $n \rightarrow \infty$, you can't use $n-1$ in the place of $m$ - it's not independent of the variable that's going to infinity.