A calculus problem by Rahil Sehgal

Since $H_n - \ln n - \gamma \; = \; \int_n^\infty \frac{x - \lfloor x\rfloor}{x^2}\,dx$ we see that $H_n - \ln n - \gamma - \tfrac{1}{2n} \; = \; \int_n^\infty \frac{x - \lfloor x \rfloor - \frac12}{x^2}\,dx \; = \; \sum_{k=n}^\infty \int_k^{k+1}\frac{x-\lfloor x \rfloor - \frac12}{x^2}\,dx \; = \; \sum_{k=n}^\infty \int_0^1 \frac{x-\frac12}{(x+k)^2}\,dx$ Since each of the integrals $\int_0^1 \frac{x-\frac12}{(x+k)^2}\,dx$ is negative, we can write \[\begin{array} {} S & = \displaystyle \; \sum_{n=1}^\infty \Big(H_n - \ln n - \gamma - \tfrac{1}{2n}\Big) \; = \; \sum_{n=1}^\infty \sum_{k=n}^\infty \int_0^1 \frac{x-\frac12}{(x+k)^2}\,dx \\ & = \displaystyle\; \sum_{k=1}^\infty \sum_{n=1}^k \int_0^1 \frac{x-\frac12}{(x+k)^2}\,dx \; =\; \sum_{k=1}^\infty k \int_0^1 \frac{x-\frac12}{(x+k)^2}\,dx \\ & = \displaystyle \sum_{k=1}^\infty k \int_0^1 \Big( \frac{1}{x+k} - \frac{k+\frac12}{(x+k)^2}\Big)\,dx \; = \; \sum_{k=1}^\infty k \,\Big[\ln(x+k) + \frac{k+\frac12}{x+k}\Big]_0^1 \\ & = \displaystyle \lim_{K \to\infty}\sum_{k=1}^K \Big(k\ln\big(\tfrac{k+1}{k}\big) - \frac{k+\frac12}{k+1}\Big) \; = \; \lim_{K \to \infty}\Big[\ln\Big(\frac{(K+1)^K}{K!}\Big) - K + \tfrac12\big(H_{K+1} - 1\big)\Big] \\ & = \displaystyle \lim_{K \to \infty}\Big[\ln\Big(\frac{(K+1)^{K+\frac12}}{e^K K!}\Big) + \tfrac12\big(H_{K+1} - \ln(K+1) - 1\big)\Big] \end{array} \] A simple application of Stirling's approximation tells us that $S \; = \; \ln\big(\tfrac{e}{\sqrt{2\pi}}\big) + \tfrac12(\gamma - 1) \; = \; \tfrac12\big(1 + \gamma - \ln(2\pi)\big)$ making the answer $1+1+1+2+2 = \boxed{7}$ .

another way is to simply write the limit $\lim_{k\to \infty} \sum_{n=1}^k (H_n-\ln(n)-\gamma-1/2n)=\lim_{k\to \infty} \left( (k+1)(H_{k+1}-1)-\ln(\Gamma(k+1))-\gamma k-\dfrac{H_k}{2}\right)$ this can be manipulated into $\lim_{k\to \infty} \left( \left(k+\dfrac{1}{2}\right) \psi(k+1)-\ln(\Gamma(k+1))-k\right)+\dfrac{\gamma}{2}$ we can use stirling approximation $\ln(\Gamma(k+1))\approx \left(k+\dfrac{1}{2}\right)\ln(k)-k+\dfrac{1}{2} \ln(2\pi)\to \left(k+\dfrac{1}{2}\right)\psi(k+1)\approx \left(k+\dfrac{1}{2}\right) \left( \dfrac{1}{2k}+\ln(k)\right) =\dfrac{1}{2}+\dfrac{1}{4k}+\left(k+\dfrac{1}{2}\right)\ln(k)$ where the second one is just the derivative of the first one multiplied by $\left(k+\dfrac{1}{2}\right)$ . plugging this in the limit we will have a lot of cancelation resulting in $\lim_{k\to \infty} \left(\dfrac{1}{2}+\dfrac{1}{4k}-\dfrac{1}{2} \ln(2\pi)\right)+\dfrac{\gamma}{2}=\boxed{\dfrac{1}{2} (1-\ln(2\pi)+\gamma)}$