# Write $a$ as a function of $n$ when $\sum_{i=1}^{n} (i + a)^{-1} = 1$

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

$$\int_a^{a+n}\frac 1{x+1}dx<\sum_{i=1}^n\frac 1{a+i}=1<\int_a^{a+n}\frac 1{x}dx\\ \biggl[\ln(x+1)\biggr]_a^{n+a}<1<\biggl[\ln (x)\biggr]_a^{a+n}\\ \ln\frac{a+n+1}{a+1}<1<\ln \frac{a+n}a\\ \ln\left(1+\frac n{a+1}\right)<1<\ln\left(1+\frac na\right)\\ 1+\frac n{a+1}<e<1+\frac na\\ \qquad \quad a<\frac n{e-1}<a+1\\ \qquad\blacksquare$$
• I also need to see how this behaves as $n \to \infty$. Any ideas? – Chris Oct 16 '15 at 1:25
• 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. – Hypergeometricx Oct 16 '15 at 1:42