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Ok, so the Euler Mascheroni constant is defined as $$\sum_{k=1}^{x} \frac1k - \ln x$$ as $x\rightarrow\infty$. However, through some fancy l'Hôpital footwork, I've discovered that the harmonic series grows at a faster rate than the natural log function, so their difference should be infinite. However, this is not the case as $\gamma$ is finite. So what gives? Thanks in advance!

P.S. Here is my footwork, I'm posting from my phone at a pizza place right now, so I didn't bother to type it all out: enter image description here

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    What fancy footwork? Doesn't seem to be valid fancy footwork, so I would post it! – 2017-02-22
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    In your development, you evaluated the derivative of $1/(x-n)$ incorrectly. You wrote $\log(x-n)$, while the actual derivative is $-1/(x-n)^2$. – 2017-02-22

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$H_n$ does not grow faster than $\log(n)$. By Frullani's theorem $$ \log\left(1+\frac{1}{n}\right) = \int_{0}^{+\infty}\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx \tag{1}$$ and since $\frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx$, $$ H_N-\log(N+1)=\sum_{n=1}^{N}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]=\int_{0}^{+\infty}\left(1-\frac{e^x-1}{xe^x}\right)\sum_{n=1}^{N}e^{-nx}\,dx \tag{2} $$ By the dominated convergence theorem, as $N\to +\infty$ the LHS of $(2)$ converges to $$ \int_{0}^{+\infty}\left(1-\frac{e^x-1}{xe^x}\right)\frac{dx}{e^x-1} = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx \tag{3}$$ that is the integral of a function in $L^1(\mathbb{R}^+)$, i.e. a finite quantity. You may also notice that the last integrand function is pretty close to $\frac{1}{2} e^{-5x/6}$, from which $\gamma\approx\frac{3}{5}$ follows.

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    Given that the OP says they are just entering a second semester calculus course, pulling out the Dominated Converge Theorem and more advanced notation like $L^1(\mathbb{R}^+)$ might be a bit much – 2017-02-22
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    Relatively heavy artillery for such a simple purpose. But I like the EM constant so much, I've posted a number of questions and answers on it. So, (+1) – 2017-02-22
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As $n \to \infty$ we have $$H_n=\ln{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots$$

Note since the first non-constant term is positive we do have that $H_n$ grows faster than $\ln(n)$ in some sense, but their growth becomes the closer to being the same as $n$ gets bigger! In other words, we have that $$\lim_{n \to \infty}\frac{H_n}{\ln(n)}=1$$

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    Oh my Gödel that's very confusing and complicated looking. I'm in Calc BC at the moment, so this is hard for me to read. Any chance you could briefly summarize this for the laymen? – 2017-02-22
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    @dsillman2000 Nothing here is really that far past a BC calc class... I'm actually in High School myself right now! What part is giving you trouble? – 2017-02-22
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    @dsillman2000 The fact you have accepted Jack's answer already concerns me considering that his uses arguably more advanced tools than my answer... are you sure you understand? – 2017-02-22
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    Well, I recognize some of the laws he was using in his answer from one of the tangents my teacher went on a week or two ago. But I've never seen anything regarding Bernoulli numbers or that strange tilde operator before. I understand most of chappers answer, though. – 2017-02-22
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    @dsillman2000 Ok, as long as you understand then all is well :) just wanted to make sure you weren't hopelessly confused. BTW, I removed the tilde operator to make what I mean a bit clearer (if less precise) – 2017-02-22
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    (+1) for showing the expansion. You might consider referencing the EMSF as a means of generating the asymptotic expansion here. -Mark – 2017-02-22
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    @Dr.MV hmm... Good idea. I might throw that up in a bit. I trimmed down my answer a TON already because the OP said it confused him/her :/ – 2017-02-22
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    Brevan. No worry. Just a suggestion. – 2017-02-22
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It is not true that $\log{n}$ grows faster than $H_n = \sum_{k=1}^n \frac{1}{k}$, or vice versa. We have $$ \frac{1}{k} \geqslant \frac{1}{x} \geqslant \frac{1}{k+1} \geqslant \frac{1}{x+1} $$ for $k \leqslant x \leqslant k+1$, and integrating from $k$ to $k+1$, $$ \frac{1}{k} \geqslant \log{(k+1)}-\log{k} \geqslant \frac{1}{k+1} \geqslant \log{(k+2)}-\log{(k+1)} $$ Thus, summing from $k=1$ to $n-1$, $$ \sum_{k=1}^{n-1} \frac{1}{k} \geqslant \log{n} - \log{1} \geqslant \sum_{k=2}^{n} \frac{1}{k} \geqslant \log{(n+2)}-\log{2}, $$ so $ H_{n-1} \geqslant \log{n} \geqslant H_{n}-1 \geqslant \log{(n+1)}-\log{2} $, i.e. $H_n$ and $\log{n}$ differ from each other only by a constant smaller than $1$.