What is $\lim\limits_{n \to \infty}\sum\limits_{k=1}^n \frac{1}{k}$ 
Possible Duplicate:
Why does 1/x diverge? 

I'm a math tutor. This is a high school level problem. I'm unable to solve this. 
What is the value of:
$\lim\limits_{n \to \infty}\sum\limits_{k=1}^n \frac{1}{k}$ 
 A: You'd think it would converge but it doesn't.
An easy way to see this is to consider the subsequences from $\frac1{2^k}$ to $\frac1{2^{k+1}}$. Since $\frac1{2^{k+1}}$ is less than $\frac1{2^k}$, consider replacing all values between them with the last one. That is:
$$1 + 1/2+1/4 + 1/4+1/8 + 1/8 + 1/8+1/8+1/8+ 1/16 + 1/16+...$$
Since each subsequence sums to $1/2$, each time you go from $n = 2^m$ to $n=2^{m+1}$, you're adding 1/2, so in the limit it won't ever converge.
A: It diverges to infinity as $\log n$.  You can see Wikipedia
A: 
From the figure, you can see that the area under the blue-curve is bounded below by the area under the red-curve from $1$ to $\infty$.
The blue-curve takes the value $\frac1{k}$ over an interval $[k,k+1)$
The red-curve is given by $f(x) = \frac1{x}$ where $x \in [1,\infty)$
The green-curve takes the value $\frac1{k+1}$ over an interval $[k,k+1)$
The area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k}$ while the area under the red-curve is given by the integral $\displaystyle \int_{1}^{n+1} \frac{dx}{x}$ while the area under the blue-curve represents the sum $\displaystyle \sum_{k=1}^{n} \frac1{k+1}$
Hence, we get $\displaystyle \sum_{k=1}^{n} \frac1{k} > \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$
$\log(n+1)$ diverges as $n \rightarrow \infty$ and hence
$$\lim_{n \rightarrow \infty} \displaystyle \sum_{k=1}^{n} \frac1{k} = + \infty$$
By a similar argument, by comparing the areas under the red curve and the green curve, we get $$\displaystyle \sum_{k=1}^{n} \frac1{k+1} < \displaystyle \int_{1}^{n+1} \frac{dx}{x} = \log(n+1)$$ and hence we can bound $\displaystyle \sum_{k=1}^{n} \frac1{k}$ from above by $1 + \log(n+1)$
Hence, $\forall n$, we have $$\log(n+1) < \displaystyle \sum_{k=1}^{n} \frac1{k} < 1 + \log(n+1)$$
Hence, we get $0 < \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1) < 1$, $\forall n$
Hence, if $a_n = \displaystyle \sum_{k=1}^{n} \frac1{k} - \log(n+1)$ we have that $a_n$ is a monotonically increasing sequence and is bounded.
Hence, $\displaystyle \lim_{n \rightarrow \infty} a_n$ exists. This limit is denoted by $\gamma$ and is called the Euler-Mascheroni constant.
It is not had to show that $\gamma \in (0.5,0.6)$ by looking at the difference in the area of these graphs and summing up the area of these approximate triangles.
