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\noindent {\bf Question}

\noindent Prove that $\frac{1}{n+1} &lt;\ln(n+1) -\ln(n)
&lt;\frac{1}{n}$ for all $n\in {\bf N}$.

\medskip
\noindent Now, consider the sequence given by $a_n =\left(
\sum_{k=1}^n \frac{1}{k} \right) -\ln(n)$. Prove that $\{ a_n\}$
is a decreasing sequence and that each $a_n$ is positive. Conclude
that the limit $\gamma =\lim_{n\rightarrow\infty} a_n$ exists.
[This number $\gamma$ is known as {\bf Euler's constant}, and
little is known about it.  For instance, it is not known whether
$\gamma$ is rational or irrational.]

\medskip

\noindent {\bf Answer}

\noindent We start with the first part of the inequality, that
$\frac{1}{n+1} &lt;\ln(n+1)-\ln(n) =\ln \left( \frac{n+1}{n}
\right)$.  Set $f(x) =\ln \left( \frac{x+1}{x} \right)
-\frac{1}{x+1}$ and $b_n =f(n)$.  We want to show that $f(x) &gt;0$
for all $x\ge 1$.  Calculating, we see that $f'(x) =
-\frac{1}{x(x+1)^2} &lt;0$ for all $x &gt;0$.  This implies that $f(x)$
is decreasing, and hence that $\{ b_n\}$ is a monotonically
decreasing sequence.  Since $\lim_{n\rightarrow\infty} b_n =0$,
this yields that $b_n &gt;0$ for all $n$.  (Because, if some $b_M
&lt;0$, then since $\{ b_n\}$ is a monotonically decreasing sequence,
we would have that $b_{M+k} &lt;b_M$ for all $k\ge 0$, and so
$\lim_{n\rightarrow\infty} b_n$ would then be negative.)   Since
$b_n &gt;0$ for all $n$, we have that $\ln \left( \frac{n+1}{n}
\right) &gt;\frac{1}{n+1}$ for all $n$, as desired.

\medskip
\noindent To handle the other part of the inequality, consider
$c_n =\frac{1}{n} - \ln \left( \frac{n+1}{n} \right)$ and set
$g(x) =\frac{1}{x} - \ln \left( \frac{x+1}{x} \right)$, so that
$c_n =g(n)$.  Since $g'(x) =-\frac{1}{x^2(x+1)}$ for all $x &gt;0$,
we see that $\{ c_n\}$ is monotonically decreasing.  Again, since
$\lim_{n\rightarrow\infty} c_n =0$, we see that $c_n &gt;0$ for all
$n$, and hence that $\frac{1}{n} &gt; \ln \left( \frac{n+1}{n}
\right)$ for all $n$, as desired.

\medskip
\noindent It remains to show that $\{ a_n\}$ is bounded below and
monotonically decreasing.  Since
\[ a_{n+1} -a_n =\left( \sum_{k=1}^{n+1} \frac{1}{k} \right) -\ln(n+1)
-\left( \sum_{k=1}^n \right) + \ln(n) = \frac{1}{n+1} -\ln(n+1)
+\ln(n) \]

$\ \ \ \ \ \ \ \ \ = \frac{1}{n+1} -\ln\left(
\frac{n+1}{n}\right), $

\noindent we see that $a_{n+1} -a_n &lt;0$ by the first part of the
inequality. That is, $\{ a_n\}$ is monotonically decreasing.

\medskip
\noindent Since $\frac{1}{n+1} &lt;\ln\left( \frac{n+1}{n}\right)$
for all $n$, we have that
\[ a_n =\left( \sum_{k=1}^n \frac{1}{k}\right) -\ln(n) = 1 + \left(
\sum_{k=1}^{n-1} \frac{1}{k+1} \right) -\ln(n) &gt; 1 +
\sum_{k=1}^{n-1} \ln \left( \frac{k+1}{k}\right) -\ln(n) = 1, \]
and so $\{ a_n\}$ is bounded below.

\medskip
\noindent Since $\{ a_n\}$ is bounded above (since $a_n &lt;a_1$ for
all $n$, since it is a monotonically decreasing sequence) and
bounded below, it is bounded.  Since it is also monotonic, we have
that $\{ a_n\}$ converges.

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