Evaluating the limit of the quotient of two infinite sums 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...

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.
