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Is there a good integral estimation technique I can apply here? Thanks!

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$$\int_a^{a+n}\frac 1{x+1}dx<\sum_{i=1}^n\frac 1{a+i}=1<\int_a^{a+n}\frac 1{x}dx\\ \biggl[\ln(x+1)\biggr]_a^{n+a}<1<\biggl[\ln (x)\biggr]_a^{a+n}\\ \ln\frac{a+n+1}{a+1}<1<\ln \frac{a+n}a\\ \ln\left(1+\frac n{a+1}\right)<1<\ln\left(1+\frac na\right)\\ 1+\frac n{a+1}<e<1+\frac na\\ \qquad \quad a<\frac n{e-1}<a+1\\ \qquad\blacksquare$$

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  • $\begingroup$ I also need to see how this behaves as $n \to \infty$. Any ideas? $\endgroup$ – Chris Oct 16 '15 at 1:25
  • $\begingroup$ What do you mean? The method provides an estimate of $a$ within a range of $1$ for any given $n$, and this range doesn't get any tighter as $n$ increases. $\endgroup$ – hypergeometric Oct 16 '15 at 1:42
  • $\begingroup$ Sorry different question. $\endgroup$ – Chris Oct 16 '15 at 1:47

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