How can I evaluate this limit?
$$\lim_{n\to\infty}\underbrace{\frac{\sum_{k=1}^n \frac 1k}{\sum_{k=1}^{n+1} \frac{1}{2k-1} }}_{=:a_n}$$
By WolframAlpha, the limit has to be 2 but how can I show this? I see it is monotonous increasing so when i could show $\sup_{n \in \mathbb N} a_n = 2$, it would be done. But I'm a bit stuck...
note that the denominator can be rewritten as $$\sum_{k=1}^{2n+1} \frac 1k - \frac 12 \sum_{k=1}^{n} \frac 1k$$ It becomes pretty easy after this: divide numerator and denominator by $\sum_{k=1}^n \frac 1k$. This gives you the limit $$\lim_{n \to \infty} \frac{1}{\frac 12 + \frac{\sum_{k=n+1}^{2n+1} \frac 1k}{\sum_{k=1}^n \frac 1k}}= 2$$
Intuitive explanation:
The ratio is
$$2\frac{\displaystyle\sum_{k=1}^n\dfrac 1k}{\displaystyle\sum_{k=1}^{n+1}\dfrac 1{k-\frac12}}$$ and for growing $k$, the term $\frac12$ becomes less and less significant. At the same time both series diverge, so that the initial terms do not matter.
By a more serious argument, you could bracket the sums by integration and obtain bounds of the form $\log n+c$. Then by squeezing
$$\frac{\log n+c_1}{\frac12\log n+c_2}<2<\frac{\log n+c_3}{\frac12\log n+c_4}.$$
L'Hopital's rule has a discrete version, under certain conditions; it is usually known as the Stolz-Cesaro theorem. Here, we treat summation as integration (and conversely, taking differences as differentiation). The statement is usually something like this: if the sequence $\{ b_n \}$ is positive and $\sum b_n = \infty$ (i.e. divergent), then for any sequence $\{ a_n \}$ of reals such that $\lim_{n\to+\infty} a_n/b_n = L$, we have
$$ \lim_{n\to+\infty} \frac{\sum_{j \le n} a_j}{\sum_{j\le n} b_j} = L. $$
A pretty cool consequence of this is the limit comparison test.
For the given example, take $a_n = 1/n$ and $b_n = 1/(2(n+1) - 1) = 1/(2n + 1)$ to get $2$ as the limit.
Comparing term by term, we have $$ \begin{align} \sum_{k=1}^n\frac1k &\le\sum_{k=1}^n\frac1{k-\frac12}\\ &=\sum_{k=1}^{n+1}\frac1{k-\frac12}-\frac1{n+\frac12} \end{align} $$ Similarly, $$ \begin{align} \sum_{k=1}^n\frac1k &\ge\sum_{k=1}^n\frac1{k+\frac12}\\ &=\sum_{k=2}^{n+1}\frac1{k-\frac12}\\ &=\sum_{k=1}^{n+1}\frac1{k-\frac12}-2 \end{align} $$ Thus, $$ 2\sum_{k=1}^{n+1}\frac1{2k-1}-2\le\sum_{k=1}^n\frac1k\le2\sum_{k=1}^{n+1}\frac1{2k-1}-\frac1{n+\frac12} $$ and therefore, $$ 2-\frac2{\sum_{k=1}^{n+1}\frac1{2k-1}}\le\frac{\sum_{k=1}^n\frac1k}{\sum_{k=1}^{n+1}\frac1{2k-1}}\le2-\frac1{\left(n+\frac12\right)\sum_{k=1}^{n+1}\frac1{2k-1}} $$ Now apply the Squeeze Theorem.
