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I see that it's an indeterminate form ($\infty-\infty$). I tried to factor and I got: $$\lim_{n\to\infty}{\frac{e^{\frac{2}{n+1}}-1}{e^{\frac{1}{n+1}}+1}\cdot e^{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}}$$

Now it's $0\cdot \infty$. How can we approach this type of problem?

Thank you!

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It may help that $\log (n+1)<H_n < 1+\log n$. – Antoine Feb 1 '14 at 12:23
up vote 6 down vote accepted


$$e^{H_{n+1}}-e^{H_n}=n(e^{1/(n+1)}-1)e^{(H_n-\ln n)}$$

If you need a further hint, try this.

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Thank you. I did the computations and I obtained $-e^\gamma$. It's correct? – Daniel C Feb 1 '14 at 13:44
@DanielC. The sign is wrong but the result is correct. – Claude Leibovici Feb 1 '14 at 13:46
@DanielC, that's what I got. (Oops, I didn't notice the negative sign. Thank you, Claude.) – Barry Cipra Feb 1 '14 at 13:46
Just to reinforce Claude's point, the harmonic sum increases with $n$, and $e^x$ is an increasing function, so the answer can't be negative. – Barry Cipra Feb 1 '14 at 13:49

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