Show that $ \lim_{n\to\infty}[1/(n+1) + 1/(n+2) + ... + 1/p*n] = \ln p $ I'm trying to refresh my math skills by working through my high school calculus text from the early 90's, Elements of Calculus and Analytic Geometry, Thomas, Finney, 1989.
At the end of Chapter 6, which introduces transcendental functions, there's a question which has me stumped:
Let $p$ be a positive integer greater than or equal to $2$.  Show that
$$
\lim_{n\to\infty}[1/(n+1) + 1/(n+2) + ... + 1/p*n] = \ln p.
$$
At first glance, the series looks like it should have a limit of 0.  I've tried breaking the limit expression into a sum of separate limits (all of which seem to go to 0), and I also tried raising e to the power of both sides.  I considered trying l'Hopital's, but that seemed like it would be a complicated mess.  I've been pondering this for more than a day now.
The problem can't be too complicated, since the book hasn't introduced anything really sophisticated, and it's just a ho-hum problem in the middle of the miscellaneous questions at the end of the chapter. 
 A: Draw the graph of $y=1/x$.  The area under this graph from $x=n$ to $x=pn$ is
$$\int_n^{pn}\frac{dx}{x}=\ln(pn)-\ln n=\ln p\ .$$
Now draw a rectangle whose base lies along the $x$ axis from $n$ to $n+1$, and whose height is $1/(n+1)$, so that it lies entirely below the curve.  Draw a similar rectangle on every unit interval up to $pn$.  The total area of these rectangles is your sum, let's call it
$$S=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{pn}\ .$$
Therefore
$$S<\ln p\ .$$
Now extend the height of the rectangles so that they all go just above the curve (so for example the first will now have height $1/n$).  The extra little rectangles you have added could all be piled up on top of each other to form a rectangle with width $1$ and height $(1/n)-(1/pn)$.  Therefore
$$S+\frac1n-\frac1{pn}>\ln p\ .$$
Putting these all together we have
$$\ln p-\frac1n+\frac1{pn}<S<\ln p\ ,$$
and now you can apply the pinching theorem (sandwich theorem, squeezing theorem).
Apologies for lack of a diagram (I'm not good at online diagrams), but if you draw one yourself I think you will find that this all becomes clear.
Comment.  As $n$ increases you have smaller and smaller terms; but you are adding up more and more of them.  This explains why the limit is not zero.
A: I wil use the following theorem:

If $x_n=1+\frac12+\cdots+\frac1n-\ln n$, then $\lim_{n\to \infty}x_n$ exists, say C. It is the Euler number.

Now 
$$\lim_{n\to\infty}[1/(n+1) + 1/(n+2) + ... + 1/p*n] \\
=\lim_{n\to\infty}[1+\frac12+\cdots+\frac{1}{pn}]-[1+\frac12+\cdots+\frac{1}{n}]\\
=\lim_{n\to \infty} [(x_{pn}+C+\ln(pn))-(x_n+C+\ln n)]
=\ln p$$
A: Writing is standard notation,
you want to show that if
$S(n, p)
=\sum_{k=n+1}^{np} \frac1{k}
$
then
$\lim_{n \to \infty} S(n, p)
=\ln p
$.
What is useful here
is the definition of $\ln$
as $\ln(x)
=\int_1^x \frac{dt}{t}
$.
This means that
$\ln(k+1)-\ln(k)
=\int_1^{k+1} \frac{dt}{t}-\int_1^k \frac{dt}{t}
=\int_k^{k+1} \frac{dt}{t}
$.
Since $\frac1{t}$
is decreasing on the interval $[k, k+1]$,
$\frac1{k+1}
\le \frac1{t}
\le \frac1{k}
$
on this interval.
Integrating these inequalities
over the interval,
since the interval is of length $1$,
$\frac1{k+1}
\le \int_k^{k+1} \frac{dt}{t}
\le \frac1{k}
$.
BTW, 
nothing about this is original on my part.
It is just the kind of thing
you will need to get comfortable with.
Therefore,
$\frac1{k+1}
\le \ln(k+1)-\ln(k)
\le \frac1{k}
$.
We are almost there now.
We now 
sum this inequality
from $k=n$ to $np-1$.
We get
$\sum_{k=n}^{np-1} \frac1{k+1}
\le \sum_{k=n}^{np-1}  (\ln(k+1)-\ln(k))
\le \sum_{k=n}^{np-1} \frac1{k}
$.
The inner sum is just
$\ln(np)-\ln(n)
=\ln(p)
$.
The left-hand sum is
$\sum_{k=n}^{np-1} \frac1{k+1}
=\sum_{k=n+1}^{np} \frac1{k}
=S(n, p)
$.
The right-hand sum is
$\sum_{k=n}^{np-1} \frac1{k}
=\sum_{k=n+1}^{np} \frac1{k}+\frac1{n}-\frac1{np}
=S(n, p)+\frac1{n}-\frac1{np}
<S(n, p)+\frac{1}{n}
$.
Therefore
$S(n, p)
\le \ln(p)
\le S(n, p)+\frac1{n}-\frac1{np}
$
or
$\ln(p)-\frac1{n}
\le S(n, p)
\le \ln(p)
$.
Letting $n \to \infty$,
we get the desired result.
