Sequences, Series and Power Series - Examples

On this page we display several examples to illustrate the concepts and theorems discussed in the Sequences, Series and Power Series theme.

The converse of Theorem 4 does not hold

Problem
Show that $\sum_{n=2}^{\infty}\frac{1}{\left(\ln(n)\right)^k}$

diverges to infinity for all $k>0$, while the limit of the sequence is $0$.

Solution

Both the root and ratio test are inconclusive for this series, so we use the comparison test. We know that $\lim_{x\rightarrow \infty}\frac{\mathrm{e}^x}{x^k}=\infty$ for every positive number $k$ (if you don't know this, use L'Hôpital repeatedly to prove it). That is, $\mathrm{e}^x$ grows faster than any polynomial. Since $\mathrm{e}^x$ is defined to be the inverse of $\ln(x)$, this implies that $\ln(x)$ grows slower than any polynomial. Thus we will eventually have, for all $n\geq N$ for some positive integer $N$,

$\ln(n)<n^{1/k}$, which implies $\left(\ln(n)\right)^k<n$. Thus

$\sum_{n=N}^\infty \frac{1}{\left(\ln(n)\right)^k}>\sum_{n=N}^\infty \frac{1}{n}=\infty.$

On the other hand, it is clear that

$\lim_{n\rightarrow \infty}\frac{1}{\left(\ln(n)\right)^k}=0$ for all $k>0$.

Estimating the sum of a series

Problem

Estimate the sums

$\text{(i)}\quad s=\sum_{n=2}^\infty \frac{\ln(n)}{n^2}$ $\text{(ii)}\quad s=\sum_{n=2}^\infty \frac{(-1)^n\ln(n)}{n^2}$

to an accuracy of $10^{-3}$

Solution
The most obvious way to estimate the sum of a series is simply to cut off the series at a certain point and calculate the resulting finite sum. However, we want to have an upper bound for the error, in order to know where to cut it off. That is, we must find a bound for the tail of series.

(i) We consider the continuous function $f(x)=\ln(x)/x^2$. It's derivative is negative for $x\geq 2$, so it is a decreasing function on the interval $[2,\infty)$. Thus, dividing this interval into sub-intervals $[n,n+1]$ of length $1$ and creating rectangles on these intervals with height equal to the value of $f$ at the left endpoints, the area of these rectangles will be greater than the area under the curve of $f$, since $f(n)$ is the greatest value of $f$ on $[n,n+1]$. Similarly, rectangles of height $f(n+1)$ will provide an under-estimate on $[n,n+1]$. Therefore,

$A_{n+1}=\int_{n+1}^\infty \frac{\ln(x)}{x^2}\,dx\leq \sum_{k=n+1}^\infty \frac{\ln(n)}{n^2}\leq \int_n^\infty \frac{\ln(x)}{x^2}\,dx=A_n.$ The integrals are easily solved and we find $\frac{\ln(n+1)+1}{n+1}\leq \sum_{k=n+1}^\infty \frac{\ln(n)}{n^2}\leq \frac{\ln(n)+1}{n}.$ Denoting the $n$th partial sum by $s_n$, we have $s\in [s_n+A_{n+1},s_n+A_n]$. We then take as our approximation the midpoint of this interval and denote it $s_n^*$. Then $|s-s_n^*|\leq \frac{A_n-A_{n+1}}{2}=\frac{\ln(n)+1}{2n}-\frac{\ln(n+1)+1}{2n+2}.$ Using maple we find that $n=43$ is the smallest for which we can guarantee that $|s-s_n^*|<10^{-3}$.

(ii) This sequence is alternating, and the absolute value of the sequence is strictly decreasing. Then $|s-s_n|<\ln(n+1)/(n+1)^2$. This is plain to see from the following argument: say $a_{n+1}$ is positive. Then in the next term, we substract a smaller number, i.e. $a_{n+1}>a_{n+1}-a_{n+2}>0$. This holds for any such pairs, so the sum of the series from the term $n+1$ cannot exceed $|a_{n+1}|$. Using maple again, we find n=64.

Using the Root Test

Problem

Does the series $\sum_{n=1}^{\infty}\left(2+\frac{1}{n}\right)^{-n}$ converge?

Solution

Because of the n'th power in the terms, we try using the root test. We have that $a_n =\left(2+\frac{1}{n}\right)^{-n}$, so all the terms are positive. We have that

$\lim_{n \to \infty} \sqrt[n]{a_n} =\lim_{n \to \infty} \sqrt[n]{\left(2+\frac{1}{n}\right)^{-n}}=\lim_{n \to \infty}\left(\frac{1}{2-1/n}\right)=\frac{1}{2}.$

Since this limit is less that 1, the root test implies that the series converge.

Finding the sum of a power series by differentiation

Problem

Find the sum and interval of convergence of the series

$\sum_{n=0}^{\infty}(n+3)x^n.$

Solution

We split the sum into two sums:

$\sum_{n=0}^{\infty}(n+3)x^n =\sum_{n=0}^{\infty}nx^n+3\sum_{n=0}^{\infty}x^n.$

The second sum is easy, we know that it is $3/(1-x)$ for $|x|<1$. For the first sum, we must do some work. We start with the identity

$1+x+x^2+ \cdots+x^n+\cdots= \sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$

which is valid for $|x|<1$.

Differentiate both sides:

$0+1+2x+ \cdots+nx^{n-1}+\cdots =\sum_{n=1}^{\infty}nx^{n-1} =\frac{1}{(1-x)^2}$

Then multiply by $x$:

$x+2x^2+ \cdots+nx^{n}+\cdots =\sum_{n=0}^{\infty}nx^{n} =\frac{x}{(1-x)^2},$

which is the sum we wanted. Note that it does not matter if we sum from $n=0$ or $n=1$, the result is the same.

In total, we have that

$\sum_{n=0}^{\infty}(n+3)x^n =\frac{x}{(1-x)^2}+\frac{3}{1-x} = \frac{3-2x}{(1-x)^2}$

for $|x|<1$. If $|x|\ge 1$, then the terms of $\sum_{n=0}^{\infty}(n+3)x^n$ do not converge to zero, so the series is not convergent in this case. Thus the interval of convergence for the series is $(-1,1)$.