I need to calculate the following sum:

$$\sum_{n=1}^\infty a_{n+1} - a_n$$

Where $$a_n = 1+\frac{1}{2}+\cdots+\frac{1}{n}- \int_1^{n+1} \frac{1}{x} \, dx$$

I have that $$a_{n+1} - a_n= \dfrac{1}{n+1} - (log(n+2)-log(n+1))$$ but I don't know what else to do, so any help would be very appreciated, thank you!

I found this in wolfram but I don't know how this was obtained:

  • $\begingroup$ Look at partial sums ($\sum_1^N$) and notice that these are indeed telescopic... $\endgroup$ – b00n heT Aug 22 '17 at 5:43

Note that the sum is telescopic: $$\sum_{n=1}^{N-1} \left(a_{n+1} - a_n\right)=a_{N}-a_1= \sum_{n=1}^N\frac{1}{n}- \int_{1}^{N+1} \dfrac{dx}{x}-1+\int_{1}^{2} \dfrac{dx}{x}\\=H_N-\ln(N+1)+\ln(2)-1$$ where $H_N=\sum_{n=1}^N\frac{1}{n}$ is the $N$-th harmonic number. Recall that, as $N$ goes to infinity, $$H_N=\ln(N)+\gamma +o(1)$$ where $\gamma$ is the Euler-Mascheroni constant. Can you take it from here?

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  • $\begingroup$ I wasn't aware of the euler-mascheroni constant which is close to the number that appears in wolfram. So what follows is to see that ln(N)- ln(N+1) converges to 0? $\endgroup$ – user392559 Aug 22 '17 at 6:22
  • $\begingroup$ @user392559 Yes, and you limit should be $\gamma+\ln(2)-1\approx 0.27036284546$. $\endgroup$ – Robert Z Aug 22 '17 at 6:58
  • $\begingroup$ Oh well, thank you! :) $\endgroup$ – user392559 Aug 22 '17 at 7:01

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