dstefan
07-01-2008, 05:17 PM
Chapter 4, Problem 6
Show that the series
\sum_{k=1}^{\infty} \frac{1}{k^2}
is convergent, while the series
\sum_{k=1}^{\infty} ~\frac{1}{k} and \sum_{k=2}^{\infty} ~\frac{1}{k \ln(k)}
are divergent.
Solution:
Since all the terms of the series \sum_{k=1}^{\infty} \frac{1}{k^2} are positive, we conclude that the series is convergent by showing that the partial sums are uniformly bounded:
\sum_{k=1}^{n} \frac{1}{k^2} = 1 + \sum_{k=1}^{n} \frac{1}{k^2} ~\leq~ 1 + \sum_{k=2}^{n} \frac{1}{k(k-1)} ~=~ 1 + \sum_{k=2}^{n} \frac{1}{k-1} - \frac{1}{k} = 1 + \left( 1 - \frac{1}{n} \right) ~<~ 2, \forall~~n \geq 2.
To show that the series \sum_{k=1}^{\infty} ~\frac{1}{k} is divergent, we prove that
\label{ineq_ex7_0}
\ln(n) ~+~ \frac{1}{n} ~<~ \sum_{k=1}^{n} \frac{1}{k} ~<~ \ln(n) ~+~ 1, \forall~~ n ~\geq~ 1
The integral of f(x) = 1/x over the interval [1,n] can be approximated from above and below as follows: Note that
\int_1^n \frac{1}{x} ~dx ~=~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx
Since f(x) = 1/x is a decreasing function, it is easy to see that \frac{1}{k+1} ~<~ f(x) ~<~ \frac{1}{k}, \forall~ x \in (k, ~k+1). Then,
\label{ineq_ex7_1}
\int_1^n \frac{1}{x} ~dx = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx ~>~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k+1} ~dx = \sum_{k=1}^{n-1} \frac{1}{k+1} = -1 ~+~ \sum_{k=1}^{n} \frac{1}{k}
\label{ineq_ex7_2}
\int_1^n \frac{1}{x} ~dx = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx ~<~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k} ~dx = \sum_{k=1}^{n-1} \frac{1}{k} = -\frac{1}{n} ~+~ \sum_{k=1}^{n} \frac{1}{k}
Note that (\ref{ineq_ex7_0}) follows from the inequalities (\ref{ineq_ex7_1}) and (\ref{ineq_ex7_2}), since
\int_1^n \frac{1}{x} ~dx ~=~ \ln(n)
In a similar fashion, by considering the integral of $\frac{1}{x \ln(x)}$
over the interval [2,n], we can show that
\label{ineq_ex7_1_second}
\ln(\ln(n)) - \ln(\ln(2)) ~+~ \frac{1}{n \ln(n)} ~<~ \sum_{k=2}^{n} ~\frac{1}{k \ln(k)}, \forall~ n \geq 2
\begin{equation}\label{ineq_ex7_2_second}
\sum_{k=2}^{n} ~\frac{1}{k \ln(k)} ~<~ \ln(\ln(n)) - \ln(\ln(2)) ~+~ \frac{1}{2 \ln(2)}, \forall~ n \geq 2
and conclude that the series \sum_{k=2}^{\infty} ~\frac{1}{k \ln(k)} is divergent.
Show that the series
\sum_{k=1}^{\infty} \frac{1}{k^2}
is convergent, while the series
\sum_{k=1}^{\infty} ~\frac{1}{k} and \sum_{k=2}^{\infty} ~\frac{1}{k \ln(k)}
are divergent.
Solution:
Since all the terms of the series \sum_{k=1}^{\infty} \frac{1}{k^2} are positive, we conclude that the series is convergent by showing that the partial sums are uniformly bounded:
\sum_{k=1}^{n} \frac{1}{k^2} = 1 + \sum_{k=1}^{n} \frac{1}{k^2} ~\leq~ 1 + \sum_{k=2}^{n} \frac{1}{k(k-1)} ~=~ 1 + \sum_{k=2}^{n} \frac{1}{k-1} - \frac{1}{k} = 1 + \left( 1 - \frac{1}{n} \right) ~<~ 2, \forall~~n \geq 2.
To show that the series \sum_{k=1}^{\infty} ~\frac{1}{k} is divergent, we prove that
\label{ineq_ex7_0}
\ln(n) ~+~ \frac{1}{n} ~<~ \sum_{k=1}^{n} \frac{1}{k} ~<~ \ln(n) ~+~ 1, \forall~~ n ~\geq~ 1
The integral of f(x) = 1/x over the interval [1,n] can be approximated from above and below as follows: Note that
\int_1^n \frac{1}{x} ~dx ~=~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx
Since f(x) = 1/x is a decreasing function, it is easy to see that \frac{1}{k+1} ~<~ f(x) ~<~ \frac{1}{k}, \forall~ x \in (k, ~k+1). Then,
\label{ineq_ex7_1}
\int_1^n \frac{1}{x} ~dx = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx ~>~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k+1} ~dx = \sum_{k=1}^{n-1} \frac{1}{k+1} = -1 ~+~ \sum_{k=1}^{n} \frac{1}{k}
\label{ineq_ex7_2}
\int_1^n \frac{1}{x} ~dx = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} ~dx ~<~ \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k} ~dx = \sum_{k=1}^{n-1} \frac{1}{k} = -\frac{1}{n} ~+~ \sum_{k=1}^{n} \frac{1}{k}
Note that (\ref{ineq_ex7_0}) follows from the inequalities (\ref{ineq_ex7_1}) and (\ref{ineq_ex7_2}), since
\int_1^n \frac{1}{x} ~dx ~=~ \ln(n)
In a similar fashion, by considering the integral of $\frac{1}{x \ln(x)}$
over the interval [2,n], we can show that
\label{ineq_ex7_1_second}
\ln(\ln(n)) - \ln(\ln(2)) ~+~ \frac{1}{n \ln(n)} ~<~ \sum_{k=2}^{n} ~\frac{1}{k \ln(k)}, \forall~ n \geq 2
\begin{equation}\label{ineq_ex7_2_second}
\sum_{k=2}^{n} ~\frac{1}{k \ln(k)} ~<~ \ln(\ln(n)) - \ln(\ln(2)) ~+~ \frac{1}{2 \ln(2)}, \forall~ n \geq 2
and conclude that the series \sum_{k=2}^{\infty} ~\frac{1}{k \ln(k)} is divergent.