Finding the smallest integer $n$ such that $1+1/2+1/3+....+1/n≥9$? I am trying to find the smallest $n$ such that $1+1/2+1/3+....+1/n \geq 9$, 
I wrote a program in C++ and found that the smallest $n$ is $4550$.
Is there any mathematical method to solve this inequality.
 A: What you are looking for is $H_n = \sum_{1 \le k \le n} k^{-1}$. It is known that $H_n \approx \ln n + \gamma - \dfrac{1}{12 n^2} + \dfrac{1}{120 n^4} - \dfrac{1}{252 n^6} + O\left(\dfrac{1}{n^8}\right)$, where $\gamma \approx 0.5772156649$ is Euler's constant.
A: We have
$$\sum_{k=1}^N \dfrac1k \sim \log(N) + \gamma + \dfrac1{N} - \dfrac1{12N^2} + \mathcal{O}(1/N^3)$$
We want the above to be equal to $9$, i.e.,
$$9 = \log(N) + \gamma + \dfrac1{N} - \dfrac1{12N^2} + \underbrace{\mathcal{O}(1/N^3)}_{\text{can be bounded by }1/N^3}$$
We have $\log(N) + \gamma < 9 \implies N \leq 4550$. Also, $\log(N) + \gamma +\dfrac1{N} > 9 \implies N \geq 4550$. Hence, $$N = 4550$$

Here is the asymptotic from Euler–Maclaurin. We have
$$\sum_{k=a}^b f(k) \sim \int_a^b f(x)dx + \dfrac{f(a) + f(b)}2 + \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)$$
In our case, we have $f(x) = \dfrac1x$, $a=1$ and $b=N$. This gives us
$$\sum_{k=1}^N \dfrac1k \sim \log(N) + \gamma + \dfrac{1}{2N} - \sum_{k=1}^{\infty} \dfrac{B_{2k}}{2k}\dfrac1{N^{2k}}$$
A: The idea was given to me by Will Jagy. I am just writing it down because he suggested I do so.
Let $H_n:=\sum_{k=1}^n\frac{1}{k}$ the $n$-th harmonic number. It is known (see here or user17762's answer above) that
$$
H_n=\ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2}\right)
$$
where $\gamma$ denotes Euler-Mascheroni constant. It is not sufficient to know that to answer your question. You need inequalities, or a precise control of the error term in the expansion. Here are two inequalities which suffice.

Claim: we have
  $$
S_n:=\ln n+\gamma <H_n< \ln n+\gamma+\frac{1}{2n}=:T_n
$$
  for all $n\geq 1$.
Application: with the lhs, a calculator, and $\gamma\simeq 0.5772156649$, we find $H_{4550}>9.00009817>9$. With the rhs, we get $H_{4549}<8.99998828<9$. So $n=4550$ is indeed the first such $n$.

Proof of the claim: first, let us show that $x_n:=H_n-S_n$ decreases. Indeed
$$
x_n-x_{n+1}=-\ln n+\ln(n+1)-\frac{1}{n+1}=-\ln\left(1-\frac{1}{n+1} \right)-\frac{1}{n+1}>0
$$
where the last inequality follows for instance from the study of $f(x)=-\ln(1-x)-x$ on $(0,1)$. Since $\lim x_n=0$, it follows that $x_n>0$ for all $n\geq 1$. That is the lhs inequality. 
Now let $y_n:=H_n-T_n$ and let us check that it increases. We have
$$
y_{n+1}-y_n=\frac{1}{2(n+1)}+\frac{1}{2n}-\ln\left(1+\frac{1}{n}\right)>0
$$
where the inequality follows from the fact that the corresponding function of $x$, instead of $n$, is decreasing on $(0,+\infty)$ with limit $0$ at $+\infty$. So $y_n$ is increasing with limit $0$. This means that $y_n<0$ for all $n\geq 1$, which proves the rhs inequality. QED.
A: In case you do not understand vonbrand and user17762's reasoning, both are using the fact that $\int_1^n \frac{dk}{k}=\ln{n}$, and that a sum can be approximated by an integral (for large n values) with a correcting factor. Here the correcting factor is $\gamma$, the Euler-Mascheroni constant.
A: You can look up "digamma function" for more info. A version I made up myself, which uses a symmetry property to give every other power, is
$$ H_n = \log \left( n + \frac{1}{2} \right) + \gamma + \frac{1}{6(2n+1)^2} - \frac{7}{60(2n+1)^4} + \frac{31}{126(2n+1)^6} - \frac{127}{120(2n+1)^8} + \frac{511}{66(2n+1)^{10}} - O \left( \frac{1}{(2n+1)^{12}} \right) $$
with
$$  \gamma = 0.5772156649015328606065... $$
I put part of this in my programmable calculator, just the 1/6 blah - 7/60 blahblah. Because of the alternating signs, this is definitive, too small with $n=4549,$ big enough with $n=4550.$ There is no guarantee the $\pm$ signs continue to alternate, and the rational coefficients are allowed to grow, note $511/66 \approx 7.74. $ The use of asymptotic expansions is to truncate at a convenient place and know enough about the possible error. Interesting to put this on a high precision program, with extremely high precision demanded for $\gamma,$ and look at the error of what I wrote multiplied by $(2n+1)^{12}.$
