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So here's a problem I've been working on for some time.

Define $\gamma$ as

$$ \gamma = 1-\int_1^\infty \frac{f(t)}{t^2} dt\,. $$

where $f(t) = t - [t]$

What I'm trying to show (unsuccessfully) is that

$$ \sum_{n \le x}\frac{1}{n} = \log x + \gamma + O(1/x) $$

I did, however, finally manage to write that in Latex :)

I believe I'll need to use partial summation, but not sure right way to do it so I get a clean answer.

Thanks for any help!

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    You can actually define $\gamma$ as $$\mathop {\lim }\limits_{n \to \infty } \left(\sum\limits_{k = 1}^n {\frac{1}{k} - \log (n+1)} \right)$$ Maybe that will help. Try to estimte the error, noting this is the difference between the integral and the rectangles in $y=x^{-1}$.2012-10-24
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    You may want to write $$\sum_{n\leq x} \frac{1}{n} = \int_{1^{-}}^{x} \frac{d\[t]}{t}$$ and then try integration by parts.2012-10-24
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    @sos440 Oh. I didn't think about Riemann-Stieltjes, just Riemann. To be more precise, if you let $${g_k} = \frac{1}{k} - \int\limits_k^{k + 1} {\frac{{dx}}{x}} $$ then $$\gamma = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {n + 1} \right)} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{g_k}} $$2012-10-24

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Abel's partial summation technique: \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} (The above is nothing but the discrete version of integration by parts).

$$\sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt$$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Consider the sum $\displaystyle \sum_{n \leq N} \frac1n$. Choose $a(n) = 1$ and $f(n) = \frac1n$. Note that we have $A(t) = \lfloor t \rfloor = t - \{t\}$. Hence, we get that \begin{align*} \sum_{n \leq N} \frac1n & = \left. \frac{t-\{t\}}t \right \rvert_{1^-}^{N^+} + \int_{1^-}^{N^+} \frac{(t-\{t\})}{t^2} dt\\ & = 1 + \int_{1^-}^{N^+} \frac{dt}t - \int_{1^-}^{N^+} \frac{\{t\}}{t^2} dt\\ & = 1 + \log (N) - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt + \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt\\ & = \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) + \log(N) + \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt \end{align*} Note that $\displaystyle \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt \leq \int_{N^+}^{\infty} \frac{1}{t^2} dt = \frac1N$.

Also note that by the same argument, we also have that $\displaystyle 0 < \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt < 1$ and hence $\displaystyle 0 < \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) < 1$. Denoting $\displaystyle \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) = \gamma$, we get the following $$ \sum_{n \leq N} \frac1n = \gamma + \log(N) + \mathcal{O} \left(\frac1N \right) $$ $\gamma \approx 0.5772\ldots$ and is called the Euler-Mascheroni constant.

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    Thanks for the reply! For some reason, I just couldn't apply the technique correctly. Nice to know I'm at least half-sane and thought it could be solved using it :) Thanks again for the response!2012-10-24
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    You know what, looking back on this result. I'm wondering actually about your notation for the integrals. What is N+ and 1- ? At first glance I thought this was just +N and -1.2012-10-24
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    To explain; this is the first time I've seen partial summation actually be represented as an integral. I knew it was the integer equivalent of integration by parts, but didn't know the two could be interchanged so easilly..2012-10-24
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    @Carl $1^-$ denotes $1- \epsilon$, where $\epsilon > 0$ is arbitrarily small. $N^+$ denotes $N + \epsilon$, where $\epsilon > 0$ is arbitrarily small.2012-10-24